Solveeit Logo
Login

Question

Question: If \[\sqrt 3 \tan \theta = 3\sin \theta \], find the value of \[{\sin ^2}\theta - {\cos ^2}\theta \]...

If 3tan⁡θ=3sin⁡θ\sqrt 3 \tan \theta = 3\sin \theta , find the value of sin⁡2θ−cos⁡2θ{\sin ^2}\theta - {\cos ^2}\theta .

Explanation

Solution

In this question, we have to find out the required trigonometric expression’s value from the given equation.
We need to first use the trigonometric formulas to bring the given equation in a shorter form so that we can find out the value of θ from the given equation then putting the value of θ in the given expression we will get the solution.
Trigonometric formula:
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}

Complete step-by-step solution:
The given trigonometric equation is 3tan⁡θ=3sin⁡θ\sqrt 3 \tan \theta = 3\sin \theta .
We need to find out the value of sin⁡2θ−cos⁡2θ{\sin ^2}\theta - {\cos ^2}\theta .
Now, we have to first find out the value of θ from the given equation.
We have,
3tan⁡θ=3sin⁡θ\sqrt 3 \tan \theta = 3\sin \theta
We know, tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.
Putting the formula in the given equation we get,
\Rightarrow$$$\sqrt 3 \dfrac{{\sin \theta }}{{\cos \theta }} = 3\sin \theta $$ By cross multiplication we get, \Rightarrow\dfrac{{\sin \theta }}{{\cos \theta }} \times \dfrac{1}{{\sin \theta }} = \dfrac{3}{{\sqrt 3 }}$$ Solving the equation we get, $\Rightarrow\dfrac{1}{{\cos \theta }} = \sqrt 3 $\Rightarrow$$$\cos \theta = \dfrac{1}{{\sqrt 3 }}
Squaring we get,
\Rightarrow$$${\cos ^2}\theta = \dfrac{1}{3}$$ We know,$${\sin ^2}\theta + {\cos ^2}\theta = 1$$ \Rightarrow{\sin ^2}\theta = 1 - {\cos ^2}\theta $$ $\Rightarrow{\sin ^2}\theta = 1 - \dfrac{1}{3} $\Rightarrow$$${\sin ^2}\theta = \dfrac{{3 - 1}}{3} = \dfrac{2}{3}
Hence, sin⁡2θ−cos⁡2θ=23−13{\sin ^2}\theta - {\cos ^2}\theta = \dfrac{2}{3} - \dfrac{1}{3}
⇒\Rightarrowsin⁡2θ−cos⁡2θ=13{\sin ^2}\theta - {\cos ^2}\theta = \dfrac{1}{3}

Hence, the value of sin⁡2θ−cos⁡2θ{\sin ^2}\theta - {\cos ^2}\theta is 13\dfrac{1}{3}.

Note: Sin Cos formulas are based on sides of the right-angled triangle. Sin and Cos are basic trigonometric functions along with tan function, in trigonometry. Sine of angle is equal to the ratio of opposite side and hypotenuse whereas cosine of an angle is equal to ratio of adjacent side and hypotenuse.
sinθ=Opposite sideHypotenuse{\text{sin}}\theta = \dfrac{{{\text{Opposite side}}}}{{{\text{Hypotenuse}}}}
cosθ=AdjacentHypotenuse{\text{cos}}\theta = \dfrac{{{\text{Adjacent}}}}{{{\text{Hypotenuse}}}}

In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. The most widely used trigonometric functions are the sine, the cosine, and the tangent.