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satela [25.4K]
3 years ago
13

[Pre-Calc]

Mathematics
1 answer:
Simora [160]3 years ago
8 0

<span><span>sin(α+β)</span>=sinαcosβ+cosαsinβ</span>,

and the double angle formula of cosine:

<span><span>cos(2α)</span>=<span>cos2α</span>−<span>sin2α</span>=2<span>cos2α</span>−1=1−2<span>sin2α</span></span>

then:

<span><span>sin(3x)</span>=<span>sin(2x+x)</span>=<span>sin(2x)</span>cosx+<span>cos(2x)</span>sinx=</span>

<span>=<span>(2sinxcosx)</span>⋅cosx+<span>(1−2<span>sin2x</span>)</span>sinx=</span>

<span>=2sinx<span>cos2x</span>+sinx−2<span>sin3x</span>=</span>

<span>=2sinx<span>(1−<span>sin2x</span>)</span>+sinx−2<span>sin3x</span>=</span>

<span>=2sinx−2<span>sin3x</span>+sinx−2<span>sin3x</span>=</span>

<span>=3sinx−4<span>sin3x</span></span>.



<span><span>cos3</span>x</span>
<span>=<span>cos<span>(x+2x)</span></span></span>
<span>=<span>cosx</span><span>cos2</span>x−<span>sinx</span><span>sin2</span>x</span><span> using cos(A+B)=cosAcosB-sinAsinB</span>
<span>=<span>cosx</span><span>(2<span><span>cos^2</span>x</span>−1)</span>−<span>sinx</span><span>(2<span>sinx</span><span>cosx</span>)</span></span><span> using cos2A=2cos^2A-1 and sin2A=2sinAcosA</span>
<span>=2<span><span>cos^3</span>x</span>−<span>cosx</span>−2<span><span>sin^2</span>x</span><span>cosx</span></span> 
<span>=2<span><span>cos^3</span>x</span>−<span>cosx</span>−2<span>(1−<span><span>cos^2</span>x</span>)</span><span>cosx</span></span><span>  using </span><span><span><span>sin^2</span>x</span>+<span><span>cos^2</span>x</span>=1</span>
<span>=2<span><span>cos^3</span>x</span>−<span>cosx</span>−2<span>(<span>cosx</span>−<span><span>cos3^</span>x</span>)</span></span><span> [open the brackets]</span>
<span>=2<span><span>cos^3</span>x</span>−<span>cosx</span>−2<span>cosx</span>+2<span><span>cos^3</span>x</span></span>
<span>=4<span><span>cos^3</span>x</span>−3<span>cosx</span></span><span> [collect terms]</span>



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