how do i solve these?

Question

1) cos(x - π/3) + sin(x - 5π/6)
looking at that confuses me big time.


and for 12tanx/1-tan^2x
how do i rewrite that one using the double angle formula?

Answer 1

1) cos(x - pi/3) + sin(x - 5pi/6)

To solve this, you may have to utilize the following identities:

cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
cos(a - b) = cos(a)cos(b) + sin(a)sin(b)
sin(a + b) = sin(a)cos(b) + sin(b)cos(a)
sin(a - b) = sin(a)cos(b) - sin(b)cos(a)

[Note, the cosine addition identity uses a minus for a plus, and a plus for a minus; the sine addition identities retain their sign).

All we have to do now is solve as per the identities, and then simplify.

cos(x - pi/3) + sin(x - 5pi/6) =
cos(x)cos(pi/3) + sin(x)sin(pi/3) + sin(x)cos(5pi/6) - sin(5pi/6)cos(x)

We know these unit circle values:
cos(pi/3) = 1/2, sin(pi/3) = sqrt(3)/2
cos(5pi/6) = sqrt(3)/2, sin(5pi/6) = 1/2

cos(x)[1/2] + sin(x)[sqrt(3)/2] + sin(x)[sqrt(3)/2] + (1/2)cos(x)

Now, we can combine like terms. The cos(x) terms will have their constants add together, as will the sin(x) terms.

(1/2) cos(x) + (1/2) cos(x) + [sqrt(3)/2]sin(x) + [sqrt(3)/2]sin(x)

[1/2 + 1/2]cos(x) + [sqrt(3)/2 + sqrt(3)/2]sin(x)
[1]cos(x) + [sqrt(3)]sin(x)
cos(x) + sqrt(3)sin(x)

2) for 12tan(x)/[1 - tan^2(x)]

Note that the double angle formula for tan goes as follows.

tan(2a) = 2tan(a) / [1 - tan^2(a)]

All we have to do is convert it to a form which utilizes this. I'll show you in steps.

12tan(x)/[1 - tan^2(x)]

I'm going to factor out a 6, and you'll see what happens.

6 [2tan(x) / [1 - tan^2(x)] ]

Now, note the double angle identity in the brackets. We can substitute this directly with tan(2x) as per that identity.

6tan(2x)

Answer 2

Use the identities for cos(x-y) and sin(x-y).
cos(x-y) = cosxcosy +sinxsiny
sin(x-y) = sinxcosy -cosxsiny
Just plug in pi/3 for y in the cos expression and 5pi/6 for y in the sin expression and get answer.

12tanx/1-tan^2x
= 6[2tanx/(1-tan^2x]
=6tan2x

Answer 3

1.cosxcospi/3+sinxsinpi/3
+sinxcos5pi/6-cosxsin5pi/6
=1/2cosx+rt3/2sinx-rt3/2sinx
-1/2cosx=0

2.2tanx/1-tan^2x=tan2x