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Section 5.2 More Trigonometric Identities
In this section, we will be working with some of these additional trigonometric identities.
Definition 5.2.1 .
The
sum and difference identities are used to find the sine or cosine of a sum or difference. These can be useful to verify other identities.
\(\displaystyle \sin(A+B)=\sin A \cos B + \cos A \sin B\)
\(\displaystyle \sin(A-B)=\sin A \cos B - \cos A \sin B\)
\(\displaystyle \cos(A+B)=\cos A \cos B - \sin A \sin B\)
\(\displaystyle \cos(A-B)=\cos A \cos B + \sin A \sin B\)
\(\displaystyle \tan(A+B)=\frac{\tan A + \tan B}{1-\tan A \tan B}\)
\(\displaystyle \tan(A-B)=\frac{\tan A - \tan B}{1+\tan A \tan B}\)
Problem 5.2.2 .
Use the sum and difference identities in
Definition 5.2.1 to prove the
double angle identities .
\(\displaystyle \sin 2A = 2\sin A \cos A\)
\(\cos 2A = \cos^2A - \sin^2A\) \(\cos 2A = 2\cos^2A - 1\) \(\cos 2A = 1 - 2 \sin^2 A\)
\(\displaystyle \tan 2A = \frac{2\tan A}{1-\tan^2 A}\)
Problem 5.2.3 .
Use the double angle identities in
Problem 5.2.2 to prove the
half angle identities .
\(\displaystyle \sin^2 \theta = \frac{1-\cos 2\theta}{2}\)
\(\displaystyle \cos^2 \theta = \frac{1+ \cos 2\theta}{2}\)
Problem 5.2.4 .
Use the sum and difference identities in
Definition 5.2.1 to verify the cofunction identities.
\(\displaystyle \sin(\frac{\pi}{2}-\theta) = \cos \theta\)
\(\displaystyle \cos(\frac{\pi}{2}-\theta) = \sin \theta\)
Problem 5.2.5 .
Use the sum and difference identities to verify the following trigonometric identities.
\(\displaystyle \sin(\theta - \pi)=-\sin \theta\)
\(\displaystyle \cos(\frac{3 \pi}{2}+\theta)=\sin \theta\)
\(\displaystyle \tan(\frac{\pi}{4}-\theta)=\frac{1-\tan \theta}{1+\tan \theta}\)
Problem 5.2.6 .
Use the identities in this section, along with any other identities you have learned to verify the following trigonometric identities.
\(\displaystyle -\frac{2\sin x \cos x}{\cos 2x}=-\tan 2x\)
\(\displaystyle \frac{\cos 2x}{2 \sin x \cos x}=\frac{1}{\tan 2x}\)
\(\displaystyle \sin x \cdot (1+ \cos 2x)=\frac{\sin 2x}{\sec x}\)
\(\displaystyle 2 \cot x \tan^2 x = \frac{\sin 2x}{\cos^2 x}\)
\(\displaystyle \frac{1- \cos 2x}{\sin 2x}=\frac{1}{\cot x}\)
