# Trigonometric Functions questions

Question 1.
Find the value of $\sqrt 3 cosec 20^0- sec 20^0$

$\sqrt 3 cosec 20^0 - sec 20^0$
$=\frac {\sqrt 3}{\sin 20} - \frac {1}{\cos 20}$
$=\frac {\sqrt 3 \cos 20 - \sin 20}{sin 20 cos 20}$
$=4\frac {(\sqrt 3/2) \cos 20 - (1/2) \sin 20}{2sin 20 cos 20}$
$=4 \frac {sin 60 cos 20 - cos60 sin 20}{sin 40}$
$=4 \frac {sin (60 0 20 )}{sin 40}$
= 4

Question 2.
Find the value of $(1 + \cos \frac {\pi}{8})(1 + \cos \frac {3\pi}{8})(1 + \cos \frac {5\pi}{8})(1 + \cos \frac {7\pi}{8})$

$(1 + \cos \frac {\pi}{8})(1 + \cos \frac {3\pi}{8})(1 + \cos \frac {5\pi}{8})(1 + \cos \frac {7\pi}{8})$
$=(1 + \cos \frac {\pi}{8})(1 + \cos \frac {3\pi}{8})(1 + \cos (\pi - \frac {3\pi}{8}))(1 + \cos ( \pi - \frac {\pi}{8}))$
$=(1 + \cos \frac {\pi}{8})(1 + \cos \frac {3\pi}{8})(1 - \cos \frac {\pi}{8})(1 - \cos \frac {3\pi}{8})$
$=(1 - \cos^2 \frac {\pi}{8})(1 - \cos^2 \frac {3\pi}{8})$
$=\sin^2 \frac {\pi}{8} \sin^2 \frac {3\pi}{8}$
Now
$\cos 2A = \cos^2 A - \sin^2 A = 1 - 2 \sin^2 A$
Therefore
$=\frac {1}{4} (1 - \cos \frac {\pi}{4})(1 - \cos \frac {3\pi}{4})$
$=\frac {1}{4} (1 - \cos \frac {\pi}{4})(1 - \cos ( \pi - \frac{\pi}{4}))$
$=\frac {1}{4} (1 - \cos \frac {\pi}{4})(1 + \cos \frac {\pi}{4})$
$=\frac {1}{4} (1 - \cos^2 \frac {\pi}{4})$
$=\frac {1}{8}$

Question 3.
If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n$, then show that $a^2+ b^2= m^2+ n^2$

$a \cos \theta + b \sin \theta = m$
Squaring both sides
$a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta=m^2$ -(1)
$a \sin \theta - b \cos \theta = n$
Squaring both the sides
$a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta=n^2$ --(2)
$a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta ) = m^2 + n^2$
or
$a^2+ b^2= m^2+ n^2$

Question 4.
if $tan (\alpha)= \frac {m}{m+1}$ and $tan (\beta)= \frac {1}{2m+1}$, Prove that $\alpha + \beta = \frac {\pi}{4}$

$tan (\alpha + \beta) = \frac {tan (\alpha) + tan (\beta)}{1 - tan (\alpha) tan (\beta)}$
Substituting the values, we have
$tan (\alpha + \beta) =1$
$tan (\alpha + \beta) = tan (\frac {\pi}{4})$
$\alpha + \beta = \frac {\pi}{4}$

Question 5.
Prove that $5 cos \theta + 3 cos(\theta + \frac {\pi}{3}) + 3$ lies between -4 and 10

$5 cos \theta + 3 cos(\theta + \frac {\pi}{3}) + 3$
$=5 cos \theta + 3 cos \theta cos \frac {\pi}{3} - 3 sin \theta sin \frac {\pi}{3} + 3$
$=\frac {1}{2} [ 13 cos \theta - 3 \sqrt 3 sin \theta ] + 3$
Let $13= r cos \alpha$ and $3 \sqrt 3 = r sin \alpha$
$r^2 =13^2 + (3 \sqrt 3)^2 = 196$
$r=14$
So equation becomes
$=\frac {1}{2} [ 14 cos (\theta + \alpha) ] + 3$
$= 7 cos (\theta + \alpha) + 3$
Now
$-1 \leq cos (\theta + \alpha) \leq 1$
$-7 \leq 7cos (\theta + \alpha) \leq 7$
$-4 \leq 7cos (\theta + \alpha) + 3 \leq 14$
Hence Proved

Question 6.
if $\frac {\pi}{2} \leq x \leq \pi$ and tan x =-4/3 , Find the value of sin (x/2),cos (x/2),tan (x/2)

$\frac {\pi}{2} \leq x \leq \pi$
$\frac {\pi}{4} \leq x/2 \leq \frac {\pi}{2}$
So, x/2 lies in First quadrant
Now As x is in second quadrant,cos x will be negative
$cos x= - \frac {1}{\sqrt {1 + tan^2x}}= - \frac {3}{5}$
Now Since x/2 lies in First quadrant, sin (x/2),cos (x/2),tan (x/2) will be positive
$cos \frac {x}{2} = \sqrt { \frac {1 + cos x}{2}}= \frac {1}{\sqrt 5}$
$sin \frac {x}{2} = \sqrt { \frac {1 - cos x}{2}}= \frac {2}{\sqrt 5}$
$tan \frac {x}{2} = \sqrt { \frac {1 - cos x}{1 + cos x}}= 2$

Question 7.
Prove that a. $\sin 12^0 \sin 48^0 \sin 54^0= \frac {1}{8}$
b. $\sin 20^0 \sin 40^0 \sin 60^0 \sin 80^0= \frac {3}{16}$

Question 8.
If $a \cos (2 \theta) + b \sin (2\theta) = c$ has $\alpha$ and $\beta$ as its roots, then prove that
$\tan (\alpha) + \tan (\beta) =\frac {2b}{a+c}$

Use the identities
$\cos (2 \theta)=\frac {1- \tan^2 \theta}{1 + \tan^2 \theta}$
and
$\sin (2 \theta)=\frac {2 \tan \theta}{1 + \tan^2 \theta}$
Substituting these values,creating quadratic in tan,then the sum of roots i.e \tan (\alpha) + \tan (\beta)$can be easily derived Question 9. Prove that$cot 7 \frac {1}{2} ^0= \sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6$Answer$cot 7 \frac {1}{2} ^0= \frac {cos 7 \frac {1}{2}^0}{sin 7 \frac {1}{2} ^0} = \frac {cos 7 \frac {1}{2}^0 cos 7 \frac {1}{2}^0}{sin 7 \frac {1}{2} ^0 cos 7 \frac {1}{2}^0}=\frac {2 cos^2 7 \frac {1}{2}^0}{2 sin 7 \frac {1}{2} ^0 cos 7 \frac {1}{2}^0}=\frac {1 + cos 15^0}{sin 15^0}=\frac {1 + cos (60 -45)}{sin (60-45)}$Now applying cos(A-B) and sin (A-B) formula and substituting the known values of 60 and 45 angles, we get$=\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6$Question 10. if$\alpha$and$\beta$are the solutions of$sin^2 x + a sin x+ b=0$as well as that of$cos^2 x + c cos (x) +d=0$,then prove that$sin (\alpha + \beta) = \frac {2ac}{a^2 + c^2}$Answer Now$sin \alpha + sin \beta = -a$-(1)$cos \alpha + cos \beta = -c$-(2) For (1)$ 2 \sin (\frac {\alpha + \beta}{2}) \cos (\frac {\alpha - \beta}{2})=-a$for (2)$ 2 \cos (\frac {\alpha + \beta}{2}) \cos (\frac {\alpha - \beta}{2})=-c$Dividing both$\tan (\frac {\alpha + \beta}{2})= \frac {a}{c}$Now$sin (\alpha + \beta)= \frac {2 tan (\frac {\alpha + \beta}{2})}{2 + tan^2 (\frac {\alpha + \beta}{2})}= \frac {2ac}{a^2 +c^2}$Question 11. If$\alpha$and$\beta$are the solutions of the equation$a tan \theta + b sec \theta = c$, then show that$tan (\alpha + \beta) = \frac {2ac}{a^2 - c^2}$Answer$a tan \theta + b sec \theta = ca tan \theta -c = -bsec \theta$squaring both the sides$(a tan \theta c)^2= b^2 sec^2 \theta(a tan \theta c)^2= b^2(1 + tan^2 \theta)$Rearranging$(a^2 b^2) tan^2 \theta 2ac tan\theta + c^2 b^2=0$Now$\alpha$and$\beta$are solutions, so$tan \alpha + tan \beta= \frac {2ac}{a^2 -b^2}tan \alpha tan \beta= \frac {c^2 -b^2}{a^2 -b^2}$Now$tan (\alpha + \beta) = \frac {tan (\alpha) + tan (\beta)}{1 - tan (\alpha) tan (\beta)}$Substituting the values, we have$tan (\alpha + \beta) =\frac {2ac}{a^2 c c^2}$Question 12. Show that$\frac {sec^2 \theta - tan \theta}{sec^2 \theta + tan \theta}$lies between 1/3 and 3 Question 13. In any$\Delta ABC$, if$a^2, b^2 ,c^2$are in A.P,Prove that cot A,cot B and Cot C are in A.P Question 14. Prove that$2 cos(x) - cos(3x) -cos (5x) = 16 cos^3 (x) sin^2 (x)$Question 15. Determine the period and Graph of$y=sin (2x)$Question 16. Determine the period and Graph of$y=-2 cos(x) + 3$Question 17. Determine the period and Graph of$y=\frac {1}{2} cos (x)$Question 18. Sketch the Graph of$y=3 sin (x)$Question 19. Prove that$\frac {cos (7x) + cos (5x)}{sin (7x) -sin (5x)}=cot (x)$Question 20. Find the value of$\tan (\frac {13 \pi}{10})\$