Question

Derive this identity from the sum and difference formulas for cosine:
cos a cos b=(1/2)[cos(a-b)+cos(a+b)]
Please show the steps to solve this

Answer 1

do you know the formula for the sum?

Answer 2

\[\overbrace{\cos(a-b)}^{\text{put the formula here}}+\overbrace{\cos(a+b)}^{\text{and formula here}}\]
add. and you will get it

Answer 3

What about this one:
Use the trigonometric subtraction formula for sine to verify this identity:
cos((pi/2)-x)=sinX
@satellite73

Answer 4

same thing, use the subtraction angle formula for cosine \[\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\] put \(a=\frac{\pi}{2},b=x\) and compute

Answer 5

Oh ok
a couple more questions:
1. If cosx=1/3 and x is in quadrant II, answer these:
a. sin(x/2)
b. tan(x/2)
2. Solve these equations graphically on the interval [0,pi]. Sketch the graph and list the solutions.
a. sinx+1=cosx
b. cos(2x)-1=sinx

How would I solve these?
@satellite73

Answer 6

ack who knows?

Answer 7

you need the half angle formula for sine for the first one

Answer 8

since you are in quadrant II, half of that angle is in quadrant 1, so sine is positive \[\sin(\frac{x}{2})=\sqrt{\frac{1-\cos(x)}{2}}\] replace \(\cos(x)\) by \(\frac{1}{3}\)

Answer 9

ok that makes sense

Answer 10

what would the equation for part b be?
@satellite73

Answer 11

i don't know the one for tangent, you should google it if it isn't in the book
can't memorize every damned thing!