I have to simplify and evaluate this :cos⁡70∘+4cos⁡70∘

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zalutaloj9a0f · Accepted Answer

The easy way :cos⁡70∘+4cos⁡70∘sin⁡70∘sin⁡70∘=sin⁡20∘+2sin⁡40∘cos⁡20∘=sin⁡20∘+sin⁡40∘+cos⁡50∘cos⁡20∘Sum to product formula gives :=cos⁡10∘+cos⁡50∘cos⁡20∘Again using the formula gives 2cos⁡30∘=3≈1.732

Leonardo Mcpherson · Accepted Answer

First 70=90−20We can express all in terms of cos⁡(20) and use that 12=cos⁡(60)=4cos3(20)−3cos⁡(20)Let&#039;s write x=cos⁡(20),y=sin⁡(20) to write less.So, 4x3−3x−12=0We square your expression such that we don&#039;t have to write radicals, but we can go without it too if we wanted.(cot⁡(70)+4cos⁡(70))2=(cos⁡(90−20)+4cos⁡(90−20)sin⁡(90−20)sin⁡(90−20))2=(y+4xyx)2=y216x2+8x+1x2But1x=8x2−6=(1−x2)(16x2+8x+1)(8x2−6)2=−1024x8−512x7+2496x6+1280x5−1952x4−1056x3+444x2+288x+36Now divide this polynomial by 4x3−3x−12, which is zero.=(−256x5−128x4+432x3+192x2−180x−66)⋅(4x3−3x−12)+3The remainder gives you the value 3.

I have to simplify and evaluate this :

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