Question

prove that sin 3 theta upon sin theta minus Cos 3 theta upon cos theta equal to 2

Answer 1

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Answer 2

Answer:

According to the given question,

We have to calculate the value of,

$\frac{sin3\theta}{sin\theta}-\frac{cos3\theta}{cos\theta}=2$

Now,

We know from the identity of the trigonometric function that,

sin3θ = 3sinθ - 4sin³θ

and,

cos3θ = 4cos³θ - 3cosθ

Now,

On putting the respective values of these functions in the given equation, we get,

$\frac{sin3\theta}{sin\theta}-\frac{cos3\theta}{cos\theta}=2\\\frac{3sin\theta-4sin^{3}\theta}{sin\theta}-\frac{4cos^{3}\theta-3cos\theta}{cos\theta}=2\\(3-4sin^{2}\theta)-(4cos^{2}\theta-3)=2\\3-4sin^{2}\theta-4cos^{2}\theta+3=2\\6-4(sin^{2}\theta+cos^{2}\theta)=2$

Now,

Also from the trigonometric identity we know that,

sin²θ + cos²θ = 1

Therefore, on putting this in the equation we get,

$6-4(sin^{2}\theta+cos^{2}\theta)=2\\6-4(1)=2\\6-4=2\\2=2$

We can see here that,

LHS = RHS

Hence, Proved.