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Question: How do you find \(\sin \) if \(\tan \) is 4?...

How do you find sin\sin if tan\tan is 4?

Explanation

Solution

In order to determine the sine when tangent of some angle xx is 4 , put tan=4\tan = 4 in the identity sec2x=tan2x+1{\sec ^2}x = {\tan ^2}x + 1 to find the value of cos2x{\cos ^2}x and put this value of cos2x{\cos^2}x in the identity of trigonometry sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1 to determine the value of sinx\sin x

Complete step by step solution:
We are given that the tangent of some xx is equal to 44
tanx=4\tan x = 4
As we know the identity of trigonometry that the sum of tangent square and one is equal to the square of secant.
sec2x=tan2x+1{\sec ^2}x = {\tan ^2}x + 1
Putting tanx=4\tan x = 4,we get
sec2x=(4)2+1 sec2x=16+1 sec2x=17 \Rightarrow {\sec ^2}x = {\left( 4 \right)^2} + 1 \\\ \Rightarrow {\sec ^2}x = 16 + 1 \\\ \Rightarrow {\sec ^2}x = 17
Secant is nothing but the reciprocal of cosine secx=1cosx\sec x = \dfrac{1}{{\cos x}}
1cos2x=17\Rightarrow \dfrac{1}{{{{\cos }^2}x}} = 17
Taking reciprocal on both sides of the equation , we get
cos2x=117\Rightarrow {\cos ^2}x = \dfrac{1}{{17}}
Now putting the above value in the identity of trigonometry sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1, we get
sin2x+117=1 sin2x=1117 sin2x=17117 sin2x=1617 \Rightarrow {\sin ^2}x + \dfrac{1}{{17}} = 1 \\\ \Rightarrow {\sin ^2}x = 1 - \dfrac{1}{{17}} \\\ \Rightarrow {\sin ^2}x = \dfrac{{17 - 1}}{{17}} \\\ \Rightarrow {\sin ^2}x = \dfrac{{16}}{{17}}
Taking square root on both sides of the equation,
sinx=1617\Rightarrow \sin x = \dfrac{{\sqrt {16} }}{{\sqrt {17} }} 16=4\sqrt {16} = 4
sinx=417\therefore \sin x = \dfrac{4}{{\sqrt {17} }}

Therefore, the value of sinx\sin x is equal to 417\dfrac{4}{{\sqrt {17} }} when tanx=4\tan x = 4.

Additional information:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2.Even Function – A function f(x)f(x) is said to be an even function ,if f(x)=f(x)f( - x) = f(x) for all x in its domain.
Odd Function – A function f(x)f(x) is said to be an even function ,if f(x)=f(x)f( - x) = - f(x) for all x in its domain.
3. We know that sin(θ)=sinθ.cos(θ)=cosθandtan(θ)=tanθ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta . Therefore,sinθ\sin \theta and tanθ\tan \theta and their reciprocals, cosecθcosec \theta and cotθ\cot \theta are odd functions whereas cosθ\cos \theta and its reciprocal secθ\sec \theta are even functions.
4. Periodic Function= A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)f(x + T) = f(x) for all x.

Note: One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.Use the identities carefully.As secant is reciprocal of cosine, also cosecant is reciprocal of sine and cotangent is reciprocal of tangent.