In other Concepts you've dealt with double angle formulas. This was useful for finding the value of an angle that was double your well known value. Now consider the idea of a "triple angle formula". If someone gave you a problem like this:
Could you compute its value?
Keep reading, and at the end of this Concept you'll know how to simplify equations such as this using the triple angle formula.
Deriving a Triple Angle Formula
Double angle formulas are great for computing the value of a trig function in certain cases. However, sometimes different multiples than two times and angle are desired. For example, it might be desirable to have three times the value of an angle to use as the argument of a trig function.
By combining the sum formula and the double angle formula, formulas for triple angles and more can be found.
Here, we take an equation which takes a linear combination of sine and cosine and converts it into a simpler cosine function.
Acosx+Bsinx=Ccos(x−D), where C=A2+B2−−−−−−−√, cosD=AC and sinD=BC.
You can also use the TI-83 to solve trigonometric equations. It is sometimes easier than solving the equation algebraically. Just be careful with the directions and make sure your final answer is in the form that is called for. You calculator cannot put radians in terms of π.
Find the formula for sin3x
Solution: Use both the double angle formula and the sum formula.
Transform 3cos2x−4sin2x into the form Ccos(2x−D)
Solution: A=3 and B=−4, so C=32+(−4)2−−−−−−−−−√=5. Therefore cosD=35 and sinD=−45 which makes the reference angle is −53.1∘ or −0.927 radians. since cosine is positive and sine is negative, the angle must be a fourth quadrant angle. D must therefore be 306.9∘ or 5.36 radians. The final answer is 3cos2x−4sin2x=5cos(2x−5.36).
Solve sinx=2cosx such that 0≤x≤2π using a graphing calculator.
Solution: In y=, graph y1=sinx and y2=2cosx.
Next, use CALC to find the intersection points of the graphs.
1. Transform 5cosx−5sinx to the form Ccos(x−D)
2. Transform −15cos3x−8sin3x to the form Ccos(x−D)
3. Derive a formula for tan4x.
1. If 5cosx−5sinx, then A=5 and B=−5. By the Pythagorean Theorem, C=52√ and cosD=552√=12√=2√2. So, because B is negative, D is in Quadrant IV. Therefore, D=7π4. Our final answer is 52√cos(x−7π4).
2. If −15cos3x−8sin3x, then A=−15 and B=−8. By the Pythagorean Theorem, C=17. Because A and B are both negative, D is in Quadrant III, which means D=cos−1(1517)=0.49+π=3.63 rad. Our final answer is 17cos3(x−3.63).
Concept Problem Solution
Using the triple angle formula we learned in this Concept for the sine function, we can break the angle down into three times a well known angle:
we can solve this problem.
Transform each expression to the form Ccos(x−D).
Derive a formula for each expression.
Find all solutions to each equation in the interval [0,2π).
Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 3.15.