How to evaluate: cos⁡24∘−cos⁡84∘−cos⁡12∘+sin⁡42∘

Question

How to evaluate:  cos⁡24∘−cos⁡84∘−cos⁡12∘+sin⁡42∘

Pansdorfp6 · Accepted Answer

cos⁡24∘−cos⁡84∘−cos⁡12∘+sin⁡42∘=cos⁡24∘−cos⁡84∘−cos⁡12∘+cos⁡48∘=cos⁡24∘+cos⁡48∘−cos⁡84∘−cos⁡12∘=2cos⁡36∘cos⁡12∘−2cos⁡48∘cos⁡36∘=2cos⁡36∘(cos⁡12∘−cos⁡48∘)=2cos⁡36∘(2sin⁡18∘sin⁡30∘)=(2sin⁡18∘cos⁡18∘)2cos⁡36∘sin⁡30∘cos⁡18∘=(2sin⁡36∘cos⁡36∘)sin⁡30∘cos⁡18∘=sin⁡72∘sin⁡30∘cos⁡18∘=sin⁡30∘=12

Vasquez · Accepted Answer

Try proving sin⁡(18∘)=5−14 Hint : If θ=18∘ Then 5θ=90∘ So 2θ=90∘−3θ ⇒sin⁡(2θ)=cos⁡(3θ) and so on. With the help of sin⁡(18∘), find sin⁡(54∘) Now , cos⁡24−cos⁡84=2sin⁡54sin⁡30cos⁡12−sin⁡42=cos⁡12−cos⁡48=2sin⁡(30)sin⁡(18) So the required value is sin⁡(54∘)−sin⁡(18∘)

alenahelenash · Accepted Answer

−cos⁡84∘−cos⁡96∘ −cos⁡12∘=cos⁡192∘=cos⁡168∘ sin⁡42∘=cos⁡(?) As cos⁡5x=−12 for x=120∘,240∘≡−120∘,480∘≡120∘,960∘≡−120∘(mod360∘) Now cos⁡5x+cos⁡x=2cos⁡3xcos⁡2x cos⁡5x=2(4cos⁡3x3cos⁡x)(2cos⁡2x1)=? So, the roots of 16c520c3+16c+12=0 are 5x=360∘n+120∘⇔x=72∘n+24∘;2≤n≤2 ∑n=22cos⁡(72∘n+24∘)=016 ⇒cos⁡(120)∘+cos⁡(48∘)+cos⁡24∘+cos⁡96∘+cos⁡168∘=0

How to evaluate: \cos 24^{\circ}-\cos 84^{\circ}-\cos 12^{\circ}+\sin 42^{\circ}

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