3.3.2: Simplifying Trigonometric Expressions using Sum and Difference Formulas
- Page ID
- 4208
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As Agent Trigonometry you are given this clue: \(\sin\left(\dfrac{\pi}{2}−x\right)\). How could you simplify this expression to make solving your case easier?
Simplifying Trigonometric Expressions
We can also use the sum and difference formulas to simplify trigonometric expressions.
The \(\sin a=−\dfrac{3}{5}\) and \(\cos b=\dfrac{12}{13}\). a is in the \(3^{rd}\) quadrant and b is in the \(1^{st}\). Let's find \(\sin(a+b)\).
First, we need to find \(\cos a\) and \(\sin b\). Using the Pythagorean Theorem, missing lengths are 4 and 5, respectively. So, \(\cos a=−\dfrac{4}{5}\) because it is in the 3rd quadrant and \(\sin b=\dfrac{5}{13}\). Now, use the appropriate formulas.
\(\begin{aligned}\sin(a+b)& =\sin a \cos b+\cos a \sin b \\&=−\dfrac{3}{5}\cdot \dfrac{12}{13}+−\dfrac{4}{5} \cdot \dfrac{5}{13}=−\dfrac{56}{65} \end{aligned}\)
Now, using the information from the previous problem above, let's find \(\tan(a+b)\).
From the cosine and sine of \(a\) and \(b\), we know that \(\tan a=\dfrac{3}{4}\) and \(\tan b=\dfrac{5}{12}\).
\(\begin{aligned} \tan(a+b)&=\dfrac{\tan a+\tan b}{1−\tan a\tan b} \\&=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}}{1−\dfrac{3}{4}\cdot \dfrac{5}{12}} \\&=\dfrac{\dfrac{14}{12}}{\dfrac{11}{16}}\\&=\dfrac{56}{33}\end{aligned}\)
Finally, let's simplify \(\cos(\pi −x)\).
Expand this using the difference formula and then simplify.
\(\begin{aligned} \cos(\pi −x)&=\cos\pi \cos x+\sin \pi \sin x\\ &=−1\cdot \cos x+0\cdot \sin x \\ &=−\cos x \end{aligned}\)
Earlier, you were asked to simplify \(\sin\left(\dfrac{\pi}{2}−x \right)\).
Solution
You can expand the expression using the difference formula and then simplify.
\(\begin{aligned} \sin\left(\dfrac{\pi}{2}−x \right)&=\sin\dfrac{\pi}{2} \cos x−\cos\dfrac{\pi}{2} \sin x \\ &=1 \cdot cosx−0\cdot \sin x \\&=\cos x \end{aligned}\)
Using the information from the first problem above (where we found \(\sin(a+b)\)), find \(\cos(a−b)\).
Solution
\(\begin{aligned} \cos(a−b)&=\cos a \cos b+\sin a \sin b=−\dfrac{4}{5} \cdot \dfrac{12}{13}+−\dfrac{3}{5} \cdot \dfrac{5}{13} \\ &=−\dfrac{63}{65} \end{aligned}\)
Simplify \(\tan(x+\pi)\).
Solution
\(\begin{aligned} \tan(x+\pi )&=\dfrac{\tan x+\tan\pi}{−\tan x \tan\pi} \\ &=\dfrac{\tan x+0}{1−\tan 0} \\&=\tan x\end{aligned}\)
Review
\(\sin a=−\dfrac{8}{17}\), \(\pi \leq a<\dfrac{3 \pi}{2}\) and \(\sin b=−\dfrac{1}{2}, \; \dfrac{3 \pi}{2}\leq b<2\pi \). Find the exact trig values of:
- \(\sin(a+b)\)
- \(\cos(a+b)\)
- \(\sin(a−b)\)
- \(\tan(a+b)\)
- \(\cos(a−b)\)
- \(\tan(a−b)\)
Simplify the following expressions.
- \(\sin(2\pi −x)\)
- \(\sin\left(\dfrac{\pi}{2}+x\right)\)
- \(\cos(x+\pi )\)
- \(\cos\left(\dfrac{3 \pi}{2}−x\right)\)
- \(\tan(x+2\pi)\)
- \(\tan(x−\pi )\)
- \(\sin\left(\pi 6−x\right)\)
- \(\tan\left(\dfrac{\pi }{4}+x\right)\)
- \(\cos\left(x−\dfrac{\pi }{3}\right)\)
Determine if the following trig statements are true or false.
- \(\sin(\pi −x)=\sin(x−\pi )\)
- \(\cos(\pi −x)=\cos(x−\pi )\)
- \(\tan(\pi −x)=\tan(x−\pi )\)
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.13.