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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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.

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