Solveeit Logo
Login

Question

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

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

Explanation

Solution

Hint:First of all, consider the given equation and use tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }. Simplify this equation to get the value of sin⁡ϕ and cos⁡ϕ\sin \phi \text{ and }\cos \phi . Now, from these values, get the corresponding sin⁡ϕ and cos⁡ϕ\sin \phi \text{ and }\cos \phi and use these two values to get value of the expression sin⁡2ϕ−cos⁡2ϕ{{\sin }^{2}}\phi -{{\cos }^{2}}\phi .

Complete step-by-step answer:
Here, we are given that 3tan⁡ϕ=3sin⁡ϕ\sqrt{3}\tan \phi =3\sin \phi . We have to find the value of sin⁡2ϕ−cos⁡2ϕ{{\sin }^{2}}\phi -{{\cos }^{2}}\phi . Let us consider the expression given in the question,
3tan⁡ϕ=3sin⁡ϕ\sqrt{3}\tan \phi =3\sin \phi
We know that, tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }. By using this in the above equation, we get,
3sin⁡ϕcos⁡ϕ=3sin⁡ϕ\sqrt{3}\dfrac{\sin \phi }{\cos \phi }=3\sin \phi
By transposing all the terms to the LHS of the above equation, we get,
3sin⁡ϕcos⁡ϕ−3sin⁡ϕ=0\sqrt{3}\dfrac{\sin \phi }{\cos \phi }-3\sin \phi =0
By taking out 3sin⁡ϕ\sqrt{3}\sin \phi common from the above equation, we get,
3sin⁡ϕ(1cos⁡ϕ−3)=0\sqrt{3}\sin \phi \left( \dfrac{1}{\cos \phi }-\sqrt{3} \right)=0
⇒3sin⁡ϕ(1−3cos⁡ϕ)=0\Rightarrow \sqrt{3}\sin \phi \left( 1-\sqrt{3}\cos \phi \right)=0
From this, we get,
3sin⁡ϕ=0;(1−3cos⁡ϕ)=0\sqrt{3}\sin \phi =0;\left( 1-\sqrt{3}\cos \phi \right)=0
sin⁡ϕ=0;3cos⁡ϕ=1\sin \phi =0;\sqrt{3}\cos \phi =1
sin⁡ϕ=0;cos⁡ϕ=13\sin \phi =0;\cos \phi =\dfrac{1}{\sqrt{3}}
Case 1: Let us take sin⁡ϕ=0\sin \phi =0
We know that,
sin⁡2ϕ+cos⁡2ϕ=1{{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1
So, we get,
0+cos⁡2ϕ=10+{{\cos }^{2}}\phi =1
cos⁡ϕ=±1\cos \phi =\pm 1
Let us consider the expression asked in the question.
E=sin⁡2ϕ−cos⁡2ϕE={{\sin }^{2}}\phi -{{\cos }^{2}}\phi
By substituting sin⁡ϕ=0\sin \phi =0 and cos⁡ϕ=±1\cos \phi =\pm 1, we get,
E=(0)2−(±1)2E={{\left( 0 \right)}^{2}}-{{\left( \pm 1 \right)}^{2}}
E=(0)2−(±1)2E={{\left( 0 \right)}^{2}}-{{\left( \pm 1 \right)}^{2}}
E=−1E = - 1
So, we get,
sin⁡2ϕ−cos⁡2ϕ=−1{{\sin }^{2}}\phi -{{\cos }^{2}}\phi =-1
Case 2: Let us take cos⁡ϕ=13\cos \phi =\dfrac{1}{\sqrt{3}}. We know that,
sin⁡2ϕ+cos⁡2ϕ=1{{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1
So, we get,
sin⁡2ϕ+(13)2=1{{\sin }^{2}}\phi +{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}=1
⇒sin⁡2ϕ+13=1\Rightarrow {{\sin }^{2}}\phi +\dfrac{1}{3}=1
⇒sin⁡2ϕ=1−13\Rightarrow {{\sin }^{2}}\phi =1-\dfrac{1}{3}
⇒sin⁡2ϕ=23\Rightarrow {{\sin }^{2}}\phi =\dfrac{2}{3}
sin⁡ϕ=±23\sin \phi =\pm \sqrt{\dfrac{2}{3}}
Let us consider the expression asked in the question,
E=sin⁡2ϕ−cos⁡2ϕE={{\sin }^{2}}\phi -{{\cos }^{2}}\phi
By substituting sin⁡ϕ=±23\sin \phi =\pm \sqrt{\dfrac{2}{3}} and cos⁡ϕ=13\cos \phi =\dfrac{1}{\sqrt{3}}, we get,
E=(±23)2−(13)2E={{\left( \pm \sqrt{\dfrac{2}{3}} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}
E=23−13E=\dfrac{2}{3}-\dfrac{1}{3}
E=13E=\dfrac{1}{3}
So, we get,
sin⁡2ϕ−cos⁡2ϕ=13{{\sin }^{2}}\phi -{{\cos }^{2}}\phi =\dfrac{1}{3}

Note: In this question, students often make this mistake of considering just the first case where we get sin⁡2ϕ−cos⁡2ϕ=−1{{\sin }^{2}}\phi -{{\cos }^{2}}\phi =-1 and leave the second case where we get, sin⁡2ϕ−cos⁡2ϕ=13{{\sin }^{2}}\phi -{{\cos }^{2}}\phi =\dfrac{1}{3}. So this must be taken care of. Since, we have got two different values of sin⁡ϕ and cos⁡ϕ\sin \phi \text{ and }\cos \phi , two different values of sin⁡2ϕ−cos⁡2ϕ{{\sin }^{2}}\phi -{{\cos }^{2}}\phi would also come.