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Published on: 16/12/2019
Trigonometric Functions
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Questions + Answers key
Take MCQ Mathematics Test
1.
prove that:
\({({sin \ 7x}+ sin 5x) +(sin 9x + sin 3x)\over (cos 7x + cos 5x) + (cos 9x + cos 3x)}=\tan6x\)
2.
Find the general solution for each of the following equations:
sec2 2x = 1- tan 2x
3.
A tree stands vertically on a hill side which makes an angle of 15\(^{o}\)with the horizontal. From a point on the ground 35m down the hill from the base of the tree, the angle of elevation of the hop of tree is 60\(^{o}\). Find the height of the tree.
4.
If \(\underset { x\longrightarrow b }{ lim } \cfrac { { x }^{ 3 }-{ b }^{ 3 } }{ x-b } =\underset { x\longrightarrow 1 }{ lim } \cfrac { { x }^{ 4 }-1 }{ x1 } \)Find all possible value of b.
5.
Find the following values: sin 36°
6.
Prove that \({sin A+sin B\over cos A+cos B}\)=\(tan ({A+B\over 2})\)
7.
Find the value of sin 75°.
8.
The perimeter of a certain sector of a circle is equal to the length of the arc of a semi-circle having the same radius. Express the angle of the sector in degrees, minutes and seconds.
9.
Find the general solutions of the following equations:
tan x + tan 2x + \(\sqrt{3}\) tan x tan 2x = \(\sqrt{3}\)
10.
Find the general solutions of the following equations \(tan \ x={-1\over \sqrt{3}}\)
11.
Find the values of tan 105o
12.
Find the values of the trigonometric function \(sin(-750^o)\)
13.
Find the degree measures corresponding to the following radian measures (use \(\pi\) = 22/7)-2c
14.
Find the degree measures corresponding to the following radian measures (use \(\pi\) = 22/7).\(({1\over6})^C\)
15.
Find the general solutions of\(\sqrt{3}\) sin x-cos x =2 ______.
X = n\(\pi +{\pi\over 6}+(-1)^n{\pi\over2}\)
X = n\(\pi -{\pi\over 6}+(-1)^n{\pi\over2}\)
X = n\(\pi +{\pi\over 6}+(-1)^{n+1}{\pi\over2}\)
None
16.
The solution of the equation cos2 \(\theta\) -+sin\(\theta\)+ 1=0lies in the interval ______.
\(({\pi\over 4},{3\pi\over4})\)
\((-{\pi\over 4},{3\pi\over4})\)
\(({3\pi\over 4},{5\pi\over4})\)
\(({5\pi\over 4},{7\pi\over4})\)
17.
In a\(\triangle\) ABC, if tan A +tanB+ tan C = o then cot A cot B cot C is equal to______.
6
3
\(1\over6\)
\(1\over3\)
18.
If tan\(\theta\) + cot \(\theta\) = 5 then tan3 \(\theta\) + cot3 \(\theta\) is equal to ______.
135
140
110
90
19.
In a ΔABC if a = 5, b = 6 and c = 5, then ∠B is ______.
cos-1 \(\left( \frac { 7 }{ 24 } \right) \)
cos-1\(\left( \frac { 7 }{ 25 } \right) \)
cos-1 \(\left( \frac { 7 }{30 } \right) \)
None
1.
We have
L.H.S. = \({({sin \ 7x}+ sin 5x) +(sin 9x + sin 3x)\over (cos 7x + cos 5x) + (cos 9x + cos 3x)}\)
\(={[2sin({7x+5x\over2})cos({7x-5x\over2})]+[2sin({9x+3x\over2})cos({9x-3x\over2})]\over[2cos({7x+5x\over2})cos({7x-5x\over2})]+[2cos({9x+3x\over2})cos({9x-3x\over2})]}\)
\(={{2sin \ 6x}cosx +2sin \ 6xcos3x \over 2cos \ 6x cosx +2cos6x cos 3x}\)
\(={{2sin \ 6x}(cosx +cos3x) \over 2cos \ 6x (cosx +cos3x )}\)
\(={sin6x\over cos6x}=\tan6x=R.H.S\).
2.
sec2 2x = 1- tan 2x
\(\Rightarrow \)1+ tan2 2x = 1- tan 2x
\(\Rightarrow \)tan2 2x + tan 2x = 0
\(\Rightarrow \)tan 2x (tan 2x + 1) = 0
\(\Rightarrow \)Either tan 2x = 0 or tan 2x + 1 = 0
\(\Rightarrow \) 2x = n\(\pi\)or tan 2x = -1
= - tan\(\pi\over4\)= tan\((-{\pi\over 4}),n \ \in \ z\)
\(\Rightarrow \) \(x={n\pi\over 2}or \ x={n\pi\over 2}+{3\pi\over 2},n\in z\)
3.
According to given information, we have the following figure.
In \(\triangle ARQ\), we have
\(\angle RAQ={ 60 }^{ \circ }and\quad \angle ARQ={ 90 }^{ \circ }\)
\(\therefore \angle AQR={ 30 }^{ \circ }\)
Now, in \(\triangle AQP\), we have \(\angle PAQ={ 45 }^{ \circ }\) and \( \angle AQP={ 30 }^{ \circ }\)
Using sine rule in \(\triangle APQ\), we get
\(\frac { AP }{ \sin { \angle AQP } } =\frac { PQ }{ \sin { \angle PAQ } } \Rightarrow PQ=35\sqrt { 2 } m\).
4.
\(\underset { x\longrightarrow b }{ lim } \cfrac { { x }^{ 3 }-{ b }^{ 3 } }{ x-b } =\underset { x\longrightarrow 1 }{ lim } \cfrac { { x }^{ 4 }-1 }{ x1 } \)
5.
We know that sin 36°= \(\sqrt{1-cos^236^o}\)
\(=\sqrt{1-({\sqrt{5}+1\over 4})^2}\)
\(={\sqrt{10-2\sqrt{5}}\over4}\)
sin 36°\(={\sqrt{10-2\sqrt{5}}\over4}\)
6.
\({sin A+sin B\over cos A+cos B}\)
\(={2sin ({A+B\over 2})cos({A-B\over2})\over 2cos({A+B\over 2})cos({A-B\over2})}\)
\(=tan ({A+B\over 2})\)
7.
sin 75°= sin (45°+ 30°)
= sin 45° cos 30° + cos 45° sin 30°
\(={1\over \sqrt{2}}\times {\sqrt{3}\over 2}+{1\over \sqrt{2}}\times {1\over2}\)
\(={\sqrt{3}\over 2\sqrt{2}}+{1\over 2\sqrt{2}}={\sqrt{3}+1\over 2\sqrt{2}}\)
8.
65° 24' 30"
9.
x =\({n\pi\over 3}+{\pi\over9},n \in z\)
10.
X = n\(\pi\) -\({\pi\over 6},\)\(n \in z\)
11.
\(-{\sqrt{3}+1\over\sqrt{3}-1}\)
12.
\(-1\over 2\)
13.
\(-({1260\over 11})\)
14.
\(({105\over 11})\)
15.
(a)
X = n\(\pi +{\pi\over 6}+(-1)^n{\pi\over2}\)
16.
(d)
\(({5\pi\over 4},{7\pi\over4})\)
17.
(c)
\(1\over6\)
18.
(c)
110
19.
(b)
cos-1\(\left( \frac { 7 }{ 25 } \right) \)
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