Simplifying trigonometric expressions

[ninatemple]

1a. sin x cos 2x + cos x sin 2x
1b. cos x cos(90-x) - sin x sin(90-x)

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my working:
1a.
sin x cos (x+x) + cos x sin (x+x)
sin x [cos x cos x - sin x sin x] + cos x [sin x cos x + cos x sin x]

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I'm not quite sure if my working is correct but I can only do halfway through.

 

[Deleted member 4993]

1a. sin x cos 2x + cos x sin 2x
1b. cos x cos(90-x) - sin x sin(90-x)

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my working:
1a.
sin x cos (x+x) + cos x sin (x+x)
sin x [cos x cos x - sin x sin x] + cos x [sin x cos x + cos x sin x]

You can go this way - but much simpler process is to realize:

sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)

In this case:

A = x & B = 2x

so:

sin x cos 2x + cos x sin 2x = sin (x+2x) = sin(3x)

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I'm not quite sure if my working is correct but I can only do halfway through.

For 1b, try to think of similar substitution as suggested above for 1a.