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Question: How do you find \( \sin \) if \( \tan = 2 \) ?...

How do you find sin⁡\sin if tan⁡=2\tan = 2 ?

Explanation

Solution

Trigonometry is used to find sides and angles. Sine, cosine and tangent are the main functions of trigonometry.

They are simply the ratio of two sides of a right-angled triangle. In the given question, we are given the tan function that is the ratio of the perpendicular and the base of a right-angled triangle, when the value of tan⁡θ\tan \theta is known, we can find out the length of the perpendicular and the base. sin⁡θ\sin \theta is the ratio of the perpendicular and the hypotenuse of the right-angled triangle so on finding out the length of the hypotenuse, we can find out the value of sin.

Complete step by step answer:
We are given that tan⁡θ=2\tan \theta = 2
And we know that
tan⁡θ=perpendicularbase ⇒21=perpendicularbase  \tan \theta = \dfrac{{perpendicular}}{{base}} \\\ \Rightarrow \dfrac{2}{1} = \dfrac{{perpendicular}}{{base}} \\\

So, perpendicular=2perpendicular = 2 and base=1base = 1

According to the Pythagoras theorem, the square of the hypotenuse is equal to the sum of the squares of the perpendicular and the base, so if the length of any two sides of the right-angled triangle is known, we can easily find out the remaining side.

So from, Pythagoras theorem, we know that –
(hypotenuse)2=(base)2+(perpendicular)2 ⇒(H)2=(2)2+(1)2 ⇒H2=4+1=5 ⇒H=±5  {(hypotenuse)^2} = {(base)^2} + {(perpendicular)^2} \\\ \Rightarrow {(H)^2} = {(2)^2} + {(1)^2} \\\ \Rightarrow {H^2} = 4 + 1 = 5 \\\ \Rightarrow H = \pm \sqrt 5 \\\

Now,
$

\sin \theta = \dfrac{{perpendicular}}{{hypotenuse}} = \dfrac{2}{{ \pm \sqrt 5 }} \\
\Rightarrow \sin \theta = \pm \dfrac{2}{{\sqrt 5 }} \\
$

Hence, if tan⁡θ=2\tan \theta = 2 then the value of sin⁡θ\sin\theta is ±25\pm \dfrac{2}{{\sqrt 5 }} .

Note: The trigonometric ratios are expressed in terms of the ratio of sides of a right-angled triangle for a specific angle θ\theta . From the value of tangent, we can find out the value of the angle and using that we can find the value of sin but that would require a calculator or vast knowledge of trigonometric functions so following the above method, we can find the value of any other trigonometric function if the value of any one trigonometric function is known.