# Using Sum, difference, and double angle identities Prove each identity using a T-chart Left Hand side Right hand sidea) cos(90degrees + A)= -sinA b)...

**Using Sum, difference, and double angle identities**

**Prove each identity** **using a T-chart**

**Left Hand side Right hand side**

**a) cos(90degrees + A)= -sinA**

**b) sin(5pi/12)=sin(pi/6)cos(pi/4)+cos(pi/6)sin(pi/4)**

**c)1+sin2x = (sinx + cosx)^2**

*print*Print*list*Cite

### 1 Answer

You need to prove that `sin((5pi)/12) = sin(pi/6)cos(pi/4)+cos(pi/6)` sin(pi/4), hence, converting `(5pi)/12 ` into the sum `(2pi)/12 +(3pi)/12` , to the left side, yields:

`sin((5pi)/12) = sin((2pi)/12 +(3pi)/12)`

Reducing duplicate factors yields:

`sin((2pi)/12 +(3pi)/12) = sin((pi)/6 +(pi)/4)`

You need to expand the sum using the following trigonometric identity, such that:

`sin(a + b) = sini a*cos b + sin b*cos a`

Reasoning by analogy yields:

`sin((pi)/6 +(pi)/4) = sin((pi)/6)cos((pi)/4) + cos((pi)/6)sin((pi)/4)`

**Notice that expanding the left side yields exactly the sum to the right side, hence, the given identity `sin((5pi)/12) = sin((pi)/6)cos((pi)/4) + cos((pi)/6)sin((pi)/4)` holds.**

You need to prove that `1+sin2x = (sinx + cosx)^2` , hence, expanding the binomial to the right side yields:

`(sinx + cosx)^2 = sin^2 x + 2 sin x*cos x + cos^2 x`

You need to remember that `sin^2 x + cos^2 x = 1` and `2 sin x*cos x` may be converted in the sine of double angle such that:

`2 sin x*cos x = sin 2x`

Hence, substituting 1 for `sin^2 x + cos^2 x` and `sin 2x` for `2 sin x*cos x` in expansion above, yields:

`(sinx + cosx)^2 = 1 + sin 2x`

**Notice that expanding the right side and making the substitutions yields exactly the expression to the left side, hence, the identity `1 + sin 2x = (sinx + cosx)^2` holds.**