Prove the trigonometric identity 9cos(a)^2-9sin(a)^2=9-18sin(a)^2

 Exercise

$9\cos^2a-9\sin^2a=9-18\sin^2a$

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Starting from the left-hand side (LHS) of the identity

$9\cos\left(a\right)^2-9\sin\left(a\right)^2$

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Applying the pythagorean identity: $\cos^2(\theta)=1-\sin(\theta)^2$

$9\left(1-\sin\left(a\right)^2\right)-9\sin\left(a\right)^2$

 Why is 1 - sin(x)^2 = cos(x)^2 ?

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Multiply the single term $9$ by each term of the polynomial $\left(1-\sin\left(a\right)^2\right)$

$9-9\sin\left(a\right)^2-9\sin\left(a\right)^2$

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Simplifying

$9-18\sin\left(a\right)^2$

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Since we have reached the expression of our goal, we have proven the identity

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