Question 257759
This you really can't solve. You can simplify or graph.
(i) {{{y = sin(4x)}}}
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Now 4x = 2x + 2x, so we get
(ii) {{{y = sin(2x + 2x)}}}
we have an identity:
sin(A+B) = sinA cosB + sinB cosA
or in our case
(iii) {{{Sin(2x+2x) = sin2xcos2x + cos2xsin2x}}}
But these are the same, so we can write
(Iv) {{{2sin(2x)}}}
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Now 2x = x + x, so we get
(v) {{{y = sin(x + x)}}}
we have an identity:
sin(A+B) = sinA cosB + sinB cosA
or in our case
(vi) {{{Sin(x+x) = sinxcosx + cosxsinx}}}
But these are the same, so we can write
(vii) {{{2sin(x)}}}
From (iv) and (vii), we get
{{{y = 4sin(x)}}}