CBSE Class 11 Maths Revision Notes Chapter 3 Trigonometric Functions

Q.1 The minute hand of a watch is 4.2 cm long. How far does its tip move in 50 minutes? (Use π = 22/7)

Ans

In 60 min, the minute hand completes 1 revolution.
∴ In 1 minute, the minute hand turns 1/60 revolution
∴ In 50 minutes, the minute hand turns
50/60 = 5/6 revolution.
Since in 1 revolution angle made by min hand is 2π radians
∴ In 5/6 revolution, angle made by min hand is
2π × 5/6 = 5π/3 radians
Here, r =4.2 cm
θ = 5π/3
⇒ Distance covered by the tip of minute hand,
l = r θ
⇒ l = 4.2 × 5π/3
⇒ l = 4.2 × 5 × (22/7) × 1/3
⇒ l = 22 cm.

Q.2

$\text{Prove that}\cos \frac{2\mathrm{\pi }}{15}·\cos \frac{4\mathrm{\pi }}{15}·\cos \frac{8\mathrm{\pi }}{15}·\cos \frac{14\mathrm{\pi }}{15}=\frac{1}{16}$

Ans

$$\begin{gathered} \text{We have} \\ \mathrm{LHS}=\cos \frac{2\mathrm{\pi }}{15}·\cos \frac{4\mathrm{\pi }}{15}·\cos \frac{8\mathrm{\pi }}{15}\cos \left(\mathrm{\pi }–\frac{\mathrm{\pi }}{15}\right) \\ =\left\{\cos \left(\frac{2\mathrm{\pi }}{15}\right)\cos \left(\frac{4\mathrm{\pi }}{15}\right)\cos \left(\frac{8\mathrm{\pi }}{15}\right)\right\}\left(–\cos \frac{\mathrm{\pi }}{15}\right) \\ =–\cos \frac{\mathrm{\pi }}{15}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{8\mathrm{\pi }}{15} \\ =–\mathrm{cosA}·\cos 2\mathrm{A}·\cos 2^2\mathrm{Acos}{2}^3{\mathrm{A}}_{\mathrm{t}}\text{where,}\mathrm{A}=\frac{\mathrm{\pi }}{15} \\ =–\left[\frac{\sin 2^4\mathrm{A}}{2^4\mathrm{sinA}}\right]=\frac{–\sin 16\mathrm{A}}{2^4\mathrm{sinA}} \\ =\frac{–\sin \left(15\mathrm{A}+\mathrm{A}\right)}{16\mathrm{sinA}}\left(\because 15\mathrm{A}=\mathrm{\pi }\right) \\ =\frac{–\sin \left(\mathrm{\pi }+\mathrm{A}\right)}{16\mathrm{sinA}}=\frac{\mathrm{sinA}}{16\mathrm{sinA}}=\frac{1}{16}=\mathrm{R}.\mathrm{H}.\mathrm{S} \end{gathered}$$

Q.3

$\text{Prove that}\sin \frac{\mathrm{\pi }}{5}\sin \frac{2\mathrm{\pi }}{5}\sin \frac{3\mathrm{\pi }}{5}\sin \frac{4\mathrm{\pi }}{5}=\frac{5}{16}$

Ans

$$\begin{gathered} \text{If}\mathrm{A}+\mathrm{B}=\mathrm{\pi },\text{then}\mathrm{A}=\mathrm{\pi }–\mathrm{B} \\ \Rightarrow \mathrm{sinA}=\sin \left(\mathrm{\pi }–\mathrm{B}\right)\Rightarrow \mathrm{sinA}=\mathrm{sinB} \\ \therefore \frac{\mathrm{\pi }}{5}+\frac{4\mathrm{\pi }}{5}=\mathrm{\pi } \\ \Rightarrow \sin \frac{\mathrm{\pi }}{5}=\sin \frac{4\mathrm{\pi }}{5}\left[\mathrm{sinA}=\mathrm{sinB}\right] \\ \text{and}\frac{2\mathrm{\pi }}{5}+\frac{3\mathrm{\pi }}{5}=\mathrm{\pi } \\ \Rightarrow \sin \frac{2\mathrm{\pi }}{5}=\sin \frac{3\mathrm{\pi }}{5}\left[\mathrm{sinA}=\mathrm{sinB}\right] \\ \text{L.H.S}=\sin \frac{\mathrm{\pi }}{5}\sin \frac{2\mathrm{\pi }}{5}\sin \frac{3\mathrm{\pi }}{5}\sin \frac{4\mathrm{\pi }}{5} \\ =\sin \frac{\mathrm{\pi }}{5}\sin \frac{2\mathrm{\pi }}{5}\sin \frac{2\mathrm{\pi }}{5}\sin \frac{\mathrm{\pi }}{5} \\ ={\left(\sin \frac{\mathrm{\pi }}{5}\sin \frac{2\mathrm{\pi }}{5}\right)}^2 \\ ={\left(\sin 36°·\sin 72°\right)}^2 \\ =\left(\sin 36°·\cos 18°\right) \\ ={\left(\frac{\sqrt{10–2\sqrt{5}}}{4}·\frac{\sqrt{10+2\sqrt{5}}}{4}\right)}^2 \\ =\frac{10–2\sqrt{5}}{16}\times \frac{10+2\sqrt{5}}{16} \\ =\frac{{\left(10\right)}^2–{\left(2\sqrt{5}\right)}^2}{256} \\ =\frac{100–20}{256}=\frac{5}{16}=\text{R.H.S.} \end{gathered}$$

Q.4 Sketch the graph of y = cos[x – (π/4)].

Ans

We have y= cos [x – (π/4)] ⇒ y – 0 = cos [x – (π/4)] …(i)
Shifting the origin at [(π/4) , 0] we obtain
x = X + (π/4) , y = Y + 0
On substituting in (i) we get Y = cos X
Now draw the graph of y = cos x and then shift it by (π/4) to the right.
The graph is shown below:

Q.5

$\text{Prove that}\frac{\cos 8\mathrm{A}\cos 5\mathrm{A}–\cos 12\mathrm{A}\cos 9\mathrm{A}}{\sin 8\mathrm{A}\cos 5\mathrm{A}+\cos 12\mathrm{A}\sin 9\mathrm{A}}=\tan 4\mathrm{A}$

Ans

$$\begin{gathered} \text{L.H.S}=\frac{2\cos 8\mathrm{A}\cos 5\mathrm{A}–2\cos 12\mathrm{A}\cos 9\mathrm{A}}{2\sin 8\mathrm{A}\cos 5\mathrm{A}+2\cos 12\mathrm{A}·\sin 9\mathrm{A}} \\ =\frac{[\cos \left(8\mathrm{A}+5\mathrm{A}\right)+\cos \left(8\mathrm{A}–5\mathrm{A}\right)]–[\cos \left(12\mathrm{A}–9\mathrm{A}\right)+\cos \left(12\mathrm{A}–9\mathrm{A}\right)]}{[\sin \left(8\mathrm{A}–5\mathrm{A}\right)+\sin \left(8\mathrm{A}–5\mathrm{A}\right)]+[\sin \left(9\mathrm{A}+12\mathrm{A}\right)+\sin \left(9\mathrm{A}–12\mathrm{A}\right)]} \\ =\frac{[\cos 13\mathrm{A}+\cos 3\mathrm{A}]–[\cos 21\mathrm{A}+\cos 3\mathrm{A}]}{[\sin 13\mathrm{A}+\sin 3\mathrm{A}]+[\sin 21\mathrm{A}+\sin (–3,\mathrm{A}]} \\ =\frac{\cos 13\mathrm{A}–\cos 21\mathrm{A}}{\sin 13\mathrm{A}+\sin 21\mathrm{A}} \\ =\frac{2\sin \left(\frac{13\mathrm{A}+21\mathrm{A}}{2}\right)\sin \left(\frac{21\mathrm{A}–13\mathrm{A}}{2}\right)}{2\sin \left(\frac{13\mathrm{A}+21\mathrm{A}}{2}\right)\cos \left(\frac{21\mathrm{A}–13\mathrm{A}}{2}\right)} \\ =\frac{\sin 17\mathrm{A}·\sin 4\mathrm{A}}{\sin 17\mathrm{A}·\cos 4\mathrm{A}}=\tan 4\mathrm{A}=\mathrm{R}·\mathrm{H}.\mathrm{S} \end{gathered}$$

Q.6

$\text{Find all other trigonometrically ratios if}\sin \mathrm{\theta }=\frac{–2\sqrt{6}}{5}\text{and}\mathrm{\theta }\text{in third quadrant.}$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }=1\Rightarrow \cos \mathrm{\theta }=\pm \sqrt{1–{\sin }^2\mathrm{\theta }} \\ \text{in the third quadrant}\cos \mathrm{\theta }\text{is negative} \\ \therefore \cos \mathrm{\theta }=–\sqrt{1–{\sin }^2\mathrm{\theta }}\Rightarrow \cos \mathrm{\theta }=–\sqrt{1–\frac{24}{25}} \\ =–\sqrt{\frac{1}{25}}=\frac{–1}{5} \\ \text{In quadrant III}\mathrm{tan\theta }\text{is positive} \\ \therefore \mathrm{tan\theta }=\frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}\Rightarrow \tan \mathrm{\theta }=\frac{–2\sqrt{6}}{5}\times \frac{–5}{1}=2\sqrt{6} \\ cosec\mathrm{\theta }=\frac{1}{\mathrm{sin\theta }}\Rightarrow cosec\mathrm{\theta }=\frac{–5}{2\sqrt{6}} \\ \sec \mathrm{\theta }=\frac{1}{\mathrm{cos\theta }}\Rightarrow \sec \mathrm{\theta }=–5 \\ \cot \mathrm{\theta }=\frac{1}{\mathrm{tan\theta }}\Rightarrow \cot \mathrm{\theta }=\frac{1}{2\sqrt{6}} \end{gathered}$$

Q.7 Prove that :

$\tan 3\mathrm{A}·\tan 2\mathrm{A}·\tan \mathrm{A}=\tan 3\mathrm{A}–\tan 2\mathrm{A}–\tan \mathrm{A}$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ 3\mathrm{A}=2\mathrm{A}+\mathrm{A} \\ \Rightarrow \tan 3\mathrm{A}=\tan \left(2\mathrm{A}+\mathrm{A}\right) \\ \Rightarrow \tan 3\mathrm{A}=\frac{\tan 2\mathrm{A}+\tan \mathrm{A}}{1–\tan 2\mathrm{A}·\tan \mathrm{A}} \\ \Rightarrow \tan 3\mathrm{A}\left(1–\tan 2\mathrm{A}·\tan \mathrm{A}\right)=\tan 2\mathrm{A}+\tan \mathrm{A} \\ \Rightarrow \tan 3\mathrm{A}–\tan 3\mathrm{A}·\tan 2\mathrm{A}·\tan \mathrm{A}=\tan 2\mathrm{A}+\tan \mathrm{A} \\ \Rightarrow \tan 3\mathrm{A}–\tan 2\mathrm{A}–\tan \mathrm{A}=\tan 3\mathrm{A}·\tan 2\mathrm{A}·\tan \mathrm{A} \end{gathered}$$

Q.8

$\mathrm{Prove}\mathrm{that}:2\cos \frac{\mathrm{\pi }}{13}\cos \frac{9\mathrm{\pi }}{13}+\cos \frac{3\mathrm{\pi }}{13}+\cos \frac{5\mathrm{\pi }}{13}=0$

Ans

$$\begin{gathered} \text{L.H.S}=2\cos \frac{\mathrm{\pi }}{13}\cos \frac{9\mathrm{\pi }}{13}+\cos \frac{3\mathrm{\pi }}{13}+\cos \frac{5\mathrm{\pi }}{13} \\ =\cos \left(\frac{9\mathrm{\pi }}{13}+\frac{\mathrm{\pi }}{13}\right)+\cos \left(\frac{9\mathrm{\pi }}{13}–\frac{\mathrm{\pi }}{13}\right)+\cos \frac{3\mathrm{\pi }}{13}+\cos \frac{5\mathrm{\pi }}{13} \\ =\cos \frac{10\mathrm{\pi }}{13}+\cos \frac{8\mathrm{\pi }}{13}+\cos \frac{3\mathrm{\pi }}{13}+\cos \frac{5\mathrm{\pi }}{13} \\ =\cos \left(\mathrm{\pi }–\frac{3\mathrm{\pi }}{13}\right)+\cos \left(\mathrm{\pi }–\frac{5\mathrm{\pi }}{13}\right)+\cos \frac{3\mathrm{\pi }}{13}+\cos \frac{5\mathrm{\pi }}{13} \\ =–\cos \frac{3\mathrm{\pi }}{13}–\cos \frac{5\mathrm{\pi }}{13}+\cos \frac{3\mathrm{\pi }}{13}+\cos \frac{5\mathrm{\pi }}{13} \\ =0=\text{R.H.S} \end{gathered}$$

Q.9

$\text{Solve}7{\cos }^2\mathrm{\theta }+3{\sin }^2\mathrm{\theta }=4.$

Ans

$$\begin{gathered} \mathrm{Since}, \\ 7{\cos }^2\mathrm{\theta }+3{\sin }^2\mathrm{\theta }=4 \\ \Rightarrow 7\left(1–{\sin }^2\mathrm{\theta }\right)+3{\sin }^2\mathrm{\theta }=4 \\ \Rightarrow 7–7{\sin }^2\mathrm{\theta }+3{\sin }^2\mathrm{\theta }=4 \\ \Rightarrow –4{\sin }^2\mathrm{\theta }=4–7 \\ \Rightarrow {\sin }^2\mathrm{\theta }=\frac{3}{4} \\ \Rightarrow {\sin }^2\mathrm{\theta }={\left(\frac{\sqrt{3}}{2}\right)}^2 \\ \Rightarrow {\sin }^2\mathrm{\theta }={\sin }^2\left(\frac{\mathrm{\pi }}{3}\right) \\ \Rightarrow \frac{1–\cos 2\mathrm{\theta }}{2}=\frac{1–\cos 2\left(\frac{\mathrm{\pi }}{3}\right)}{2} \\ \Rightarrow \cos 2\mathrm{\theta }=\cos 2\left(\frac{\mathrm{\pi }}{3}\right) \\ \Rightarrow 2\mathrm{\theta }=2\mathrm{n\pi }\pm \frac{2\mathrm{\pi }}{3} \\ \Rightarrow \mathrm{\theta }=\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z} \\ \text{Thus, the general solution of given equation is}\mathrm{\theta }=\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z}\text{.} \end{gathered}$$

Q.10

$\mathrm{Show}\mathrm{that}:\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\mathrm{\theta }}}}=2\mathrm{cos\theta }$

Ans

$$\begin{gathered} \mathrm{L}·\mathrm{H}.\mathrm{S}=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 8\mathrm{\theta })}}} \\ =\sqrt{2\sqrt{2+\sqrt{\left(2\times 2{\cos }^{2}4\mathrm{\theta }\right)}}}\left(\because 1+\cos 8\mathrm{\theta }=2{\cos }^{2}4\mathrm{\theta }\right) \\ =\sqrt{2+\sqrt{2+\sqrt{4{\cos }^{2}4\mathrm{\theta }}}} \\ =\sqrt{2+\sqrt{2+2\cos 4\mathrm{\theta }}} \\ =\sqrt{2+\sqrt{2\left(1+\cos 4\mathrm{\theta }\right)}} \\ =\sqrt{2+\sqrt{2\times 2{\cos }^{2}2\mathrm{\theta }}}\left\{\because 1+\cos 4\mathrm{\theta }=2{\cos }^{2}2\mathrm{\theta }\right\} \\ =\sqrt{2+\sqrt{4{\cos }^{2}2\mathrm{\theta }}} \\ =\sqrt{2+2\cos 2\mathrm{\theta }} \\ =\sqrt{2\left(1+\cos 2\mathrm{\theta }\right)} \\ =\sqrt{2.2{\cos }^2\mathrm{\theta }} \\ =2\mathrm{cos\theta }=\mathrm{R}·\mathrm{H}.\mathrm{S} \end{gathered}$$

Q.11

$\text{Prove that :}\tan 6°\tan 42°\tan 66°\tan 78°=1$

Ans

$$\begin{gathered} =\frac{\sin 6°\sin 42°\sin 66°\sin 78°}{\cos 6°\cos 42°\cos 66°\cos 78°} \\ =\frac{\left(2\sin 66°\sin 6°\right)\left(2\sin 78°\sin 42°\right)}{\left(2\cos 66°\cos 6°\right)\left(2\cos 78°\cos 42°\right)} \\ =\frac{\left(\cos 60°–\cos 72°\right)\left(\cos 36°–\cos 120°\right)}{\left(\cos 60°+\cos 72°\right)\left(\cos 36°+\cos 120°\right)} \\ =\frac{\left(\frac{1}{2}–\frac{\sqrt{5}–1}{4}\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right)}{\left(\frac{1}{2}+\frac{\sqrt{5}–1}{4}\right)\left(\frac{\sqrt{5}+1}{4}–\frac{1}{2}\right)} \\ =\frac{\left(\frac{2–\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}+1+2}{4}\right)}{\left(\frac{2+\sqrt{5}–1}{4}\right)\left(\frac{\sqrt{5}+1–2}{4}\right)} \\ =\frac{\left(3–\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}–1\right)} \\ =\frac{9–5}{5–1} \\ =\frac{4}{4}=1=\mathrm{RHS} \end{gathered}$$

Q.12

$\text{Prove that}\cos 510°·\cos 330°+\sin 390°·\cos 120°=–1$

Ans

$$\begin{gathered} \text{L.H.S.}=\cos 510°\cos 330°+\sin 390°\cos 120° \\ =\cos \left(360°+150°\right)\cos \left(360°–30°\right)+\sin \left(360°+30°\right)\cos \left(180°+60°\right) \\ =\cos 150°\cos 30°+\sin 30°\times –\cos 60° \\ =\cos \left(180°–30°\right)\cos 30°–\sin 30°\cos 60° \\ =–\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}–\frac{1}{2}\times \frac{1}{2} \\ =–\frac{3}{4}–\frac{1}{4} \\ =–1 \\ =\text{R.H.S.} \end{gathered}$$

Q.13 Find the degree measure corresponding to the radian measure (2π/15).

Ans

We have π radian = 180^°
1 radian =(180 /π)°

$\frac{2\mathrm{\pi }}{15}\text{radian}=\left(\frac{2\mathrm{\pi }}{15}\times \frac{180}{\mathrm{\pi }}\right)°=24°$

Q.14 If three angles A, B, C, are in A.P. Prove that:

$\mathrm{cotB}=\frac{\mathrm{sinA}–\mathrm{sinC}}{\mathrm{cosC}–\mathrm{cosA}}$

Ans

$$\begin{gathered} \mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{\mathrm{sinA}–\mathrm{sinC}}{\mathrm{cosC}–\mathrm{cosA}} \\ =\frac{2\sin \frac{\mathrm{A}–\mathrm{C}}{2}\cos \frac{\mathrm{A}+\mathrm{C}}{2}}{2\sin \frac{\mathrm{A}+\mathrm{C}}{2}\sin \frac{\mathrm{A}–\mathrm{C}}{2}} \\ =\cot \left(\frac{\mathrm{A}+\mathrm{C}}{2}\right)=\mathrm{cotB}=\mathrm{L}.\mathrm{H}.\mathrm{S} \\ \left[\begin{array}{c}\because \mathrm{A},\mathrm{B},\mathrm{C}\text{are in A.P}\\ \therefore 2\mathrm{B}=\mathrm{A}+\mathrm{C}\end{array}\right] \end{gathered}$$

Q.15

$\text{Prove that}\frac{\sin 2\mathrm{\theta }}{1+\cos 2\mathrm{\theta }}=\mathrm{tan\theta }$

Ans

$$\begin{gathered} \text{L.H.S}=\frac{\sin 2\mathrm{\theta }}{1+\cos 2\mathrm{\theta }}=\frac{2·\mathrm{sin\theta }·\mathrm{cos\theta }}{2{\cos }^2\mathrm{\theta }}=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }} \\ =\mathrm{tan\theta }=\text{R.H.S} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}·\mathrm{H}.\mathrm{S} \end{gathered}$$

Q.16 (π/8) radian = ……degree.

Ans

(π/8) radian

$=\left(\frac{\mathrm{\pi }}{8}\times \frac{180}{\mathrm{\pi }}\right)°=\left(\frac{45}{2}\right)°=22\frac{1}{2}°=22°{\left(\frac{1}{2}\times 60\right)}^{‘}=22°{30}^{‘}$

Q.17 Find solution of cos x = (1/2).

Ans

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{cosx}=\frac{1}{2} \\ \mathrm{cosx}=\cos \left(\frac{\mathrm{\pi }}{3}\right) \\ \text{Thus,}\mathrm{x}=2\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3} \\ \mathrm{where}\mathrm{n}\mathrm{is}\mathrm{any}\mathrm{integer}. \end{gathered}$$

Q.18

–37°30′ is = …… radian

Ans

$$\begin{gathered} –37°{30}^{‘}=–\left(37\frac{1}{2}\right)° \\ =–\left(\frac{75}{2}\right)° \\ =–\left(\frac{75}{2}\times \frac{\mathrm{\pi }}{180}\right)\text{radian} \\ =–\left(\frac{5\mathrm{\pi }}{24}\right)\text{radian} \end{gathered}$$

Q.19 Find solution of sin x = (√3/2).

Ans

$$\begin{gathered} \mathrm{Here}, \\ \mathrm{sinx}=–\frac{\sqrt{3}}{2} \\ \mathrm{sinx}=–\sin \left(\frac{\mathrm{\pi }}{3}\right) \\ =\sin \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{3}\right) \\ =\sin \left(\frac{4\mathrm{\pi }}{3}\right) \\ \text{Thus,}\mathrm{x}=\mathrm{n\pi }+{\left(–1\right)}^{\mathrm{n}}\frac{4\mathrm{\pi }}{3} \\ \mathrm{where}\mathrm{n}\mathrm{is}\mathrm{any}\mathrm{integer}. \end{gathered}$$

Q.20 If cos x = –(1/2), x lies in third quadrant find sin x and cot x.

Ans

$$\begin{gathered} \mathrm{sinx}=–\sqrt{1–{\cos }^2},\left(\because \mathrm{x}\mathrm{lies}\mathrm{in}\mathrm{third}quadrant\right) \\ =–\sqrt{1–{\left(–\frac{1}{2}\right)}^2}=–\sqrt{\frac{3}{4}}=–\frac{\sqrt{3}}{2} \\ \cot \mathrm{x}=\sqrt{{cosec}^2\mathrm{x}–1} \\ =\sqrt{\frac{1}{{\sin }^2\mathrm{x}}–1} \\ =\sqrt{\frac{4}{3}–1}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}} \end{gathered}$$

Q.21 Show that

$\tan 3\mathrm{x}=\frac{3\mathrm{tanx}–{\tan }^3\mathrm{x}}{1–3{\tan }^2\mathrm{x}}$

Ans

$$\begin{gathered} \tan 3\mathrm{x}=\tan (2\mathrm{x}+\mathrm{x}) \\ =\frac{\tan 2\mathrm{x}+\tan \mathrm{x}}{1–\tan 2\mathrm{x}·\tan \mathrm{x}} \\ =\frac{\frac{2\tan \mathrm{x}}{1–{\tan }^2\mathrm{x}}+\tan \mathrm{x}}{1–\frac{2\tan \mathrm{x}}{1–{\tan }^2\mathrm{x}}·\tan \mathrm{x}} \\ =\frac{3\tan \mathrm{x}–{\tan }^3\mathrm{x}}{1–3{\tan }^2\mathrm{x}} \end{gathered}$$

Q.22 Find the value of sin15°.

Ans

$$\begin{gathered} \sin 15°=\sin \left(45°–30°\right) \\ =\sin 45°\cos 30°–\cos 45°\sin 30° \\ =\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}–\frac{1}{\sqrt{2}}\frac{1}{2}=\frac{\sqrt{3}–1}{2\sqrt{2}} \end{gathered}$$

Q.23 Find the value of

${\sin }^4\sqrt{\frac{\mathrm{\pi }}{6}}+2{\sin }^2\sqrt{\frac{\mathrm{\pi }}{6}}{\cos }^2\sqrt{\frac{\mathrm{\pi }}{6}}+{\cos }^4\sqrt{\frac{\mathrm{\pi }}{6}}$

.

Ans

$$\begin{gathered} {\sin }^4\sqrt{\frac{\mathrm{\pi }}{6}}+2{\sin }^2\sqrt{\frac{\mathrm{\pi }}{6}}{\cos }^2\sqrt{\frac{\mathrm{\pi }}{6}}+{\cos }^4\sqrt{\frac{\mathrm{\pi }}{6}} \\ ={\left({\sin }^2\sqrt{\frac{\mathrm{\pi }}{6}}\right)}^2+2{Sin}^2\sqrt{\frac{\mathrm{\pi }}{6}}{\cos }^2\sqrt{\frac{\mathrm{\pi }}{6}}+{\left({\cos }^2\sqrt{\frac{\mathrm{\pi }}{6}}\right)}^2 \\ ={\left({\sin }^2\sqrt{\frac{\mathrm{\pi }}{6}}+{\cos }^2\sqrt{\frac{\mathrm{\pi }}{6}}\right)}^2 \\ =1^2=1 \end{gathered}$$

Q.24 If tan (π/6) = (1/√3), then find the value of tan [π – (π/6)].

Ans

tan [π – (π/6)] = – tan(π/6) = – (1/√3)

Q.25 State sine rule.

Ans

If the sides of the triangle are a, b and c and the angles opposite those
sides are A, B and C respectively, then the law of sine states :

$\frac{\mathrm{a}}{\mathrm{sinA}}=\frac{\mathrm{b}}{\mathrm{sinB}}=\frac{\mathrm{c}}{\mathrm{sinC}}$

Q.26 Find the Value of cos (13π/12) .

Ans

$$\begin{gathered} \cos \frac{13\mathrm{\pi }}{12}=\cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{12}\right)=–\cos \frac{\mathrm{\pi }}{12} \\ –\cos \left(\frac{\mathrm{\pi }}{4}–\frac{\mathrm{\pi }}{6}\right)=–\left[\cos \frac{\mathrm{\pi }}{4}\cos \frac{\mathrm{\pi }}{6}+\sin \frac{\mathrm{\pi }}{4}\sin \frac{\mathrm{\pi }}{6}\right] \\ –\left[\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}\right]=–\left[\frac{\sqrt{3}+1}{2\sqrt{2}}\right] \end{gathered}$$

Q.27 State the laws of cosine.

Ans

The laws of cosine are:

$$\begin{gathered} {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2–2\mathrm{ab}\mathrm{cosC} \\ \text{or, equivalently:} \\ {\mathrm{a}}^2={\mathrm{b}}^2+{\mathrm{c}}^2–2\mathrm{bc}\mathrm{cosA} \\ {\mathrm{b}}^2={\mathrm{c}}^2+{\mathrm{a}}^2–2\mathrm{ca}\mathrm{cosB} \end{gathered}$$

Q.28 If the arcs of the same length in two circles subtend angles 65°and 110° at the centre, then find the ratio of their radii.

Ans

$$\begin{gathered} \text{Let}{\mathrm{r}}_1\text{and}{\mathrm{r}}_2\text{be the radii of the two circles.} \\ \text{Given that} \\ {\mathrm{\theta }}_1=45°=\frac{\mathrm{\pi }}{180°}\times 45°=\frac{\mathrm{\pi }}{4}\text{radian} \\ {\mathrm{\theta }}_2=100°=\frac{\mathrm{\pi }}{180°}\times 100°=\frac{5\mathrm{\pi }}{9}\text{radian} \\ \text{Let I be the length of each of the arc. Then,} \\ \mathrm{I}={\mathrm{\theta }}_1{\mathrm{r}}_1={\mathrm{\theta }}_2{\mathrm{r}}_2 \\ \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{{\mathrm{\theta }}_2}{{\mathrm{\theta }}_1} \\ \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{\frac{5\mathrm{\pi }}{9}}{\frac{\mathrm{\pi }}{4}}=\frac{5}{9}\times 4 \\ \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{20}{9} \\ \text{Hence,}{\mathrm{r}}_1:{\mathrm{r}}_2=20:9 \end{gathered}$$

Q.29 In a circle of diameter 42 cm, the length of a chord is 21 cm. Find the length of the minor arc of the chord.

Ans

$$\begin{gathered} \text{Diameter}=42\mathrm{cm} \\ \therefore \text{Radius,}\mathrm{r}=\frac{42}{2}=21\mathrm{cm} \\ \text{Chord,}\mathrm{AB}=21\mathrm{cm} \\ \Rightarrow ∆\mathrm{AOB}\text{is equilateral} \\ \therefore \angle \mathrm{AOB}=60°=\frac{\mathrm{\pi }}{180°}\times 60°=\frac{\mathrm{\pi }}{3}\mathrm{radian} \\ \Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{3}\mathrm{radian},\mathrm{r}=21\mathrm{cm} \\ \therefore \text{The length of the minor arc,} \\ \mathrm{l}=\mathrm{\theta }\mathrm{x}\mathrm{r} \\ =\frac{\mathrm{\pi }}{3}\times 21 \\ =\frac{22}{7}\times \frac{1}{3}\times 21\mathrm{cm} \\ =22\mathrm{cm} \end{gathered}$$

Q.30 If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, then find the ratio of their radii.

Ans

$$\begin{gathered} \text{Let}{\mathrm{r}}_1\text{and}{\mathrm{r}}_2\text{be the radii of the two circles.} \\ \text{Given that} \\ {\mathrm{\theta }}_1=60°=\frac{\mathrm{\pi }}{180°}\times 60°=\frac{\mathrm{\pi }}{3}\text{radian} \\ {\mathrm{\theta }}_2=75°=\frac{\mathrm{\pi }}{180°}\times 75°=\frac{5\mathrm{\pi }}{12}\text{radian} \\ \text{Let I be the length of each of the arc. Then,} \\ \mathrm{l}={\mathrm{\theta }}_1{\mathrm{r}}_1={\mathrm{\theta }}_2{\mathrm{r}}_2 \\ \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{{\mathrm{\theta }}_2}{{\mathrm{\theta }}_1} \\ \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{\frac{\mathrm{\pi }}{3}}{\frac{5\mathrm{\pi }}{12}}=\frac{1}{3}\times \frac{12}{5} \\ \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{4}{5} \\ \text{Hence,}{\mathrm{r}}_1:{\mathrm{r}}_2=4:5 \end{gathered}$$

Q.31

$\text{Prove that}\cos 2\mathrm{x}=\frac{1–{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}}$

Ans

$$\begin{gathered} \text{We know that}\cos \left(\mathrm{x}+\mathrm{y}\right)=\cos \mathrm{x}\cos \mathrm{y}–\sin \mathrm{x}\sin \mathrm{y} \\ \text{Replacing}\mathrm{y}\text{by}\mathrm{x},\text{we get} \\ \cos \left(\mathrm{x}+\mathrm{x}\right)=\cos \mathrm{x}\cos \mathrm{x}–\sin \mathrm{x}\sin \mathrm{x} \\ \Rightarrow \cos 2\mathrm{x}={\cos }^2\mathrm{x}–{\sin }^2\mathrm{x} \\ =\frac{{\cos }^2\mathrm{x}–{\sin }^2\mathrm{x}}{1} \\ =\frac{{\cos }^2\mathrm{x}–{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}} \\ \text{Divide Numerator and Denominator by}{\cos }^2\mathrm{x} \\ =\frac{1–\frac{{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}}{1+\frac{{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}} \\ =\frac{1–{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}} \end{gathered}$$

Q.32

$\text{If}\sqrt{7}\tan \mathrm{x}=1,\text{then find the value of}\frac{{cosec}^2\mathrm{x}–{\sec }^2\mathrm{x}}{{cosec}^2\mathrm{x}+{\sec }^2\mathrm{x}}$

Ans

$$\begin{gathered} \because \sqrt{7}\tan \mathrm{x}=1 \\ \therefore \tan \mathrm{x}=\frac{1}{\sqrt{7}} \\ \text{The value of =}\frac{{cosec}^2\mathrm{x}–{\sec }^2\mathrm{x}}{{cosec}^2\mathrm{x}+{\sec }^2\mathrm{x}} \\ =\frac{\frac{1}{{\sin }^2\mathrm{x}}–\frac{1}{{\cos }^2\mathrm{x}}}{\frac{1}{{\sin }^2\mathrm{x}}+\frac{1}{{\cos }^2\mathrm{x}}} \\ =\frac{{\cos }^2\mathrm{x}–{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}} \\ \text{Divide Numerator and Denominator by}{\cos }^2\mathrm{x} \\ =\frac{1–\frac{{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}}{1+\frac{{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}} \\ =\frac{1–{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}} \\ =\frac{1–\frac{1}{7}}{1+\frac{1}{7}} \\ =\frac{\frac{6}{7}}{\frac{8}{7}}=\frac{6}{8}=\frac{3}{4} \end{gathered}$$

Q.33

$\text{Prove that}\sin 3\mathrm{x}=3\mathrm{sinx}–4{\sin }^3\mathrm{x}$

Ans

$$\begin{gathered} \text{LHS}=\sin 3\mathrm{x} \\ =\sin \left(2\mathrm{x}+\mathrm{x}\right) \\ =\sin 2\mathrm{x}\cos \mathrm{x}+\cos 2\mathrm{x}\sin \mathrm{x} \\ =2\sin \mathrm{x}\cos \mathrm{x}\cos \mathrm{x}+\left(1–2{\sin }^2\mathrm{x}\right)\sin \mathrm{x} \\ =2\sin \mathrm{x}\left(1–{\sin }^2\mathrm{x}\right)+\sin \mathrm{x}–2{\sin }^3\mathrm{x} \\ =2\sin \mathrm{x}\left(1–{\sin }^3\mathrm{x}\right)+\sin \mathrm{x}–2{\sin }^3\mathrm{x} \\ =3\sin \mathrm{x}–4{\sin }^3\mathrm{x} \end{gathered}$$

Q.34

$\text{Prove that}\mathrm{tanx}=\sqrt{\frac{1–\cos 2\mathrm{x}}{1+\cos 2\mathrm{x}}}.$

Ans

$$\begin{gathered} \mathrm{RHS}=\sqrt{\frac{1–\cos 2\mathrm{x}}{1+\cos 2\mathrm{x}}} \\ =\sqrt{\frac{1–\left(1–2{\sin }^2\mathrm{x}\right)}{1+2{\cos }^2\mathrm{x}–1}} \\ =\sqrt{\frac{1–1+2{\sin }^2\mathrm{x}}{2{\cos }^2\mathrm{x}}} \\ =\sqrt{\frac{2{\sin }^2\mathrm{x}}{2{\cos }^2\mathrm{x}}} \\ =\sqrt{\frac{{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}} \\ =\frac{\mathrm{sinx}}{\mathrm{cosx}} \\ =\mathrm{tanx} \\ \text{= LHS, Proved} \end{gathered}$$

Q.35

$\text{Evaluate}\tan \left(22\frac{1°}{2}\right)\text{.}$

Ans

$$\begin{gathered} \text{We know that} \\ \tan \mathrm{x}=\sqrt{\frac{1–\cos 2\mathrm{x}}{1+\cos 2\mathrm{x}}}...\left(\mathrm{i}\right) \\ \text{Now put}\mathrm{x}=22\frac{1°}{2}=\frac{45°}{2}\text{in}\left(1\right)\text{we get,} \\ \tan 22\frac{1°}{2}=\sqrt{\frac{1–\cos \left(2\times \frac{45°}{2}\right)}{1+\cos \left(2\times \frac{45°}{2}\right)}} \\ \tan \frac{45°}{2}=\sqrt{\frac{1–\cos 45°}{1+\cos 45°}} \\ =\sqrt{\frac{1–\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}} \\ =\sqrt{\frac{\sqrt{2}–1}{\sqrt{2}+1}} \\ =\sqrt{\frac{\left(\sqrt{2}–1\right)}{\left(\sqrt{2}–1\right)}\times \frac{\left(\sqrt{2}–1\right)}{\left(\sqrt{2}–1\right)}} \\ =\sqrt{\frac{{\left(\sqrt{2}–1\right)}^2}{2–1}}=\left(\sqrt{2}–1\right) \end{gathered}$$

Q.36 Which is greater – sin1° or sin1? Justify your answer.

Ans

$$\begin{gathered} \text{First, we shall convert}1\text{into degree} \\ \because \mathrm{\pi }=180° \\ \therefore 1=\left(\frac{180}{\mathrm{\pi }}\right)° \\ =\left(\frac{180}{\frac{22}{7}}\right)° \\ =\left(180\times \frac{7}{22}\right)° \\ =\left(90\times \frac{7}{11}\right)° \\ =\left(\frac{630}{11}\right)° \\ \Rightarrow 1=57.27° \\ \Rightarrow \sin 1=\sin 57.27° \\ \text{Hence,}\sin 1\text{is greater than}\sin 1°\text{.} \end{gathered}$$

Q.37

$\text{Prove that}\tan \frac{3\mathrm{x}}{2}=\frac{3\tan \frac{\mathrm{x}}{2}–{\tan }^3\frac{\mathrm{x}}{2}}{1–3{\tan }^2\frac{\mathrm{x}}{2}}$

Ans

$$\begin{gathered} \mathrm{LHS}=\tan \frac{3\mathrm{x}}{2} \\ =\tan \left(\mathrm{x}+\frac{\mathrm{x}}{2}\right) \\ =\frac{\tan \mathrm{x}+\tan \frac{\mathrm{x}}{2}}{1–\tan \mathrm{x}\tan \frac{\mathrm{x}}{2}} \\ =\frac{\left(\frac{2\tan \frac{\mathrm{x}}{2}}{1–{\tan }^2\frac{\mathrm{x}}{2}}\right)+\tan \frac{\mathrm{x}}{2}}{1–\left(\frac{2\tan \frac{\mathrm{x}}{2}}{1–{\tan }^2\frac{\mathrm{x}}{2}}\right)\tan \frac{\mathrm{x}}{2}} \\ =\frac{\frac{2\tan \frac{\mathrm{x}}{2}+\tan \frac{\mathrm{x}}{2}–{\tan }^3\frac{\mathrm{x}}{2}}{\left(1–{\tan }^2\frac{\mathrm{x}}{2}\right)}}{\frac{1–{\tan }^2\frac{\mathrm{x}}{2}–2{\tan }^2\frac{\mathrm{x}}{2}}{\left(1–{\tan }^2\frac{\mathrm{x}}{2}\right)}} \\ =\frac{3\tan \frac{\mathrm{x}}{2}–{\tan }^3\frac{\mathrm{x}}{2}}{1–3{\tan }^2\frac{\mathrm{x}}{2}} \\ =\text{RHS, Proved.} \end{gathered}$$

Q.38 Evaluate sin18°.

Ans

Q.39 Evaluate cos72°.

Ans

$$\begin{gathered} \text{Let}\mathrm{\theta }=18° \\ \Rightarrow 5\mathrm{\theta }=90° \\ \Rightarrow 3\mathrm{\theta }+2\mathrm{\theta }=90° \\ \Rightarrow 2\mathrm{\theta }=90°–3\mathrm{\theta } \\ \Rightarrow \sin 2\mathrm{\theta }=\sin \left(90°–3\mathrm{\theta }\right) \\ \Rightarrow \sin 2\mathrm{\theta }=\cos 3\mathrm{\theta } \\ \Rightarrow \cos 3\mathrm{\theta }=\sin 2\mathrm{\theta } \\ \Rightarrow 4{\cos }^3\mathrm{\theta }–3\mathrm{cos\theta }=2\sin \mathrm{\theta }\cos \mathrm{\theta } \\ \Rightarrow 4{\cos }^3\mathrm{\theta }–3\mathrm{cos\theta }–2\sin \mathrm{\theta }\cos \mathrm{\theta }=0 \\ \Rightarrow \cos \mathrm{\theta }\left[4{\cos }^2\mathrm{\theta }–3–2\sin \mathrm{\theta }\right]=0 \\ \Rightarrow \cos \mathrm{\theta }\left[4\left(1–{\sin }^2\mathrm{\theta }\right)–3–2\sin \mathrm{\theta }\right]=0 \\ \Rightarrow \cos \mathrm{\theta }\left[4–4{\sin }^2\mathrm{\theta }–3–2\sin \mathrm{\theta }\right]=0 \\ \Rightarrow \cos \mathrm{\theta }\left[1–4{\sin }^2\mathrm{\theta }–2\sin \mathrm{\theta }\right]=0 \\ \Rightarrow –\cos \mathrm{\theta }\left[4{\sin }^2\mathrm{\theta }+2\sin \mathrm{\theta }–1\right]=0 \\ \Rightarrow \cos \mathrm{\theta }\left[4{\sin }^2\mathrm{\theta }+2\sin \mathrm{\theta }–1\right]=0 \\ \Rightarrow \cos \mathrm{\theta }\left[4{\sin }^2\mathrm{\theta }+2\sin \mathrm{\theta }–1\right]=0 \\ \because \cos \mathrm{\theta }\ne 0\text{or}\mathrm{\theta }\ne 90° \\ \Rightarrow 4{\sin }^2\mathrm{\theta }+2\sin \mathrm{\theta }–1=0 \\ \text{This is a quadratic equation in}\sin \mathrm{\theta } \\ \text{with}\mathrm{a}=4,\mathrm{b}=2\text{and}\mathrm{c}=–1 \\ \therefore \mathrm{D}={\mathrm{b}}^2–4\mathrm{ac} \\ =4–4\times 4\times –1 \\ =4+16 \\ =20 \\ \Rightarrow \sin \mathrm{\theta }=\frac{–\mathrm{b}\pm \sqrt{\mathrm{D}}}{2\mathrm{a}}=\frac{–2\pm \sqrt{20}}{2\times 4} \\ \Rightarrow \sin \mathrm{\theta }=\frac{–2\pm 2\sqrt{5}}{2\times 4}=\frac{–1\pm \sqrt{5}}{4} \\ \Rightarrow \sin \mathrm{\theta }=\frac{–1+\sqrt{5}}{4}\text{and} \\ \sin \mathrm{\theta }=\frac{–1–\sqrt{5}}{4}\left[\mathrm{rejected}\mathrm{as}\mathrm{in}\mathrm{I}\mathrm{quadrant}\mathrm{all}+\mathrm{ve}\right] \\ \sin 18°=\frac{\sqrt{5}–1}{4} \\ \cos 72°=\cos \left(90°–18°\right) \\ \Rightarrow \cos 72°=\sin 18° \\ \Rightarrow \cos 72°=\frac{\sqrt{5}–1}{4} \end{gathered}$$