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I'm doing practice problems out of Trigonometry 10th ed. Hornsby and ran into a question.

Section 5.1 question 71:

Write each expression in terms of sine and cosine, and simplify so that no quotients appear in the final expression and all functions are of $\theta$.

$\frac{1-\cos^2(-\theta)}{1+\tan^2(-\theta)}$

The book has the answer $\sin^2 \theta \cos^2 \theta$. However, I cannot figure out how to get to this answer.

I started by pulling out the negative from $\theta$:

$\frac{1+\cos^2\theta}{1-\tan^2\theta}$

Then I changed tangent into sine and cosine: $\frac{1+\cos^2\theta}{1-\frac{\sin\theta}{\cos\theta}}$

Then multiplied by the reciprocal:

$(1+\cos^2\theta)(1-\frac{\cos\theta}{\sin\theta})$

And this is where things got really confusing as I had no idea what I could do with the result of the above expression.

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When you pull the negative out of $\theta$, nothing changes since the terms with it are squared. We see $$\frac{1-\cos^2(-\theta)}{1+\tan^2(-\theta)} = \frac{1- \cos^2(\theta)}{1 + \tan^2(\theta)} = \frac{\sin^2(\theta))}{1+\tan^2(\theta)} = \left(\frac{\cos^2(\theta)}{\cos^2(\theta)}\right)\frac{\sin^2(\theta))}{1+\tan^2(\theta)}.$$ Now $\cos^2(\theta) \tan^2(\theta) = \sin^2(\theta)$ so we get $$\frac{1-\cos^2(-\theta)}{1+\tan^2(-\theta)} = \frac{\cos^2(\theta) \sin^2(\theta)}{\cos^2(\theta) + \sin^2(\theta)} = \sin^2(\theta)\cos^2(\theta).$$

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  • $\begingroup$ Thank you, the tip about the negative $\theta$ solved the problem for me! $\endgroup$ – Michael McQuade Mar 28 '16 at 4:30
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HINT:

$$\tan^2(-\theta)=\{\tan(-\theta)\}^2=(-\tan\theta)^2=\tan^2\theta$$

$$\cos^2(-\theta)=\{\cos(-\theta)\}^2=(\cos\theta)^2=\cos^2\theta$$

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HINT: Check that $$ 1+\tan^2 x=\frac{1}{\cos^2x}. $$ And use Pythagorean trigonometric identity.

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