Solving Trig Identities Practice Problems

Solving Trig Identities Practice Problems-63
Note how we work on one side only and pull down the other side when it matches.

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Sometimes we have to find common denominators, like in the last example.

We didn’t need to turn it into sin and cos, since we only had tan and cot in the identity (although it still would have worked).

I memorize the \(A-A\) part and then remember that I can use the Pythagorean Identity \(A A=1\) to substitute and do the algebra to arrive at the other expressions.

For \(\tan \left( \right)\), the identity only has tan’s in it, with the “These are a little more complicated.

You’ve already seen the reciprocal and quotient identities.

You can also write these as “\(\sin x\)”, and so on.\(\displaystyle \begin\cos \left( \right)\cos \left( \right)&=\frac\\left( \right)\left( \right)&=\frac\\left( \right)\left( \right)&=\frac\\fracx-\fracx&=\frac\\frac\left( \right)-\fracx&=\frac\3-3x-x&=2\-4x&=-1\x&=\frac\\sin x&=\pm \frac\end\) \(\displaystyle x=\left\) We use double angle and half angle identities the same way we used sum and difference identities when we need to split up the angle to make it easier to find the values (for example, to find values on the unit circle).We also use the identities in conjunction with other identities to prove and solve trig problems.Here are some examples of simple identity proofs with reciprocal and quotient identities.Typically, to do these proofs, you must always start with one side (either side, but usually take the more complicated side) and manipulate the side until you end up with the other side.\(\displaystyle \begin&=\frac=\frac=\frac\&=\frac\cdot \frac\&=\frac\&=\frac\\frac&=\frac\,\,\,\,\,\,\,\,\surd \end\) Note: The right-hand side looked a little more complicated (because of the tangents) so we started there.We turned the tangents into sines and cosines and simplified first.Dividing In this problem, it is easier to start from the RHS. We multiply numerator and denominator of the fraction by the conjugate of the denominator.\(\displaystyle \sin \theta =\frac\,\,\,\,\,\,\,\,\,\,\,\csc \theta =\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \theta =\frac\,\) \(\displaystyle \tan \theta =\frac=\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \theta =\frac=\frac\) \(\begin\sin \left( \right)=\sin A\cos B \cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( \right)=\cos A\cos B-\sin A\sin B\\sin \left( \right)=\sin A\cos B-\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( \right)=\cos A\cos B \sin A\sin B\end\) \(\displaystyle \theta \theta =1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta 1=\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta 1=\theta \) An “identity” is something that is always true, so you are typically either substituting or trying to get two sides of an equation to equal each other.Then we multiplied by \(\displaystyle \begin\frac&=\\frac&=\\frac\cdot \frac&=\\frac&=\\frac&=\frac\,\,\,\,\,\,\surd \end\) Note: We knew to use the \(\displaystyle 2A-1\) for \(\cos 2A\) because of the cos in the numerator; then we could factor.We multiplied by Since \(\displaystyle \cos \left( \right)=\pm \sqrt\), we need to get \(\cos A\), and then find the quadrant of \(\displaystyle \cos \left( \right)\) to get the correct sign.

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