# How to verify this trigonometric identity?

$9\cos^2 B − 9\sin^2 B = 18\cos^2 B − 9$

Thanks

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$9\cos^2(B)-9\sin^2(B)=9\cos^2(B)-9(1-\cos^2(B))=18\cos^2(B)-9$

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Recall that $$\sin^2(B) = 1-\cos^2(B)$$ Plug this into the left hand side to get what you want.

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Since $\cos^2 B$ appears in two places, you could start by bringing everything to one side of the equation to see whether you get something nicer. In this case you get

$$9\cos^2 B+9\sin^2 B-9=0\;,$$

which is true if and only if $\cos^2B+\sin^2B-1=0$, i.e., if and only if $\cos^2B+\sin^2B=1$, which you know is true. Now start with the known $\cos^2B+\sin^2B=1$ and work back through the calculations to prove the identity.

This is not the most efficient approach, but it’s a good way to discover a proof if you don’t happen to see one of the more efficient calculations right away.

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\begin{align*} \text{LHS} &= 9\cos^2 B − 9\sin^2 B \\ &= 9(\cos^2 B-\sin^2 B) \\ &= 9(\cos^2 B - (1-\cos^2 B)) \\ &= 9(2\cos^2 B- 1) \\ &= 18\cos B - 9 \\ &= \text{RHS}. \end{align*}

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