Trigonometry problem sin⁡100∘+cos⁡70∘cos⁡80∘−cos⁡20∘I've done trigonometry in my earlier years of high school but I forgot a lot of rules. This is where I'm stuck on this problem:sin⁡100∘+cos⁡70∘cos⁡80∘−cos⁡20∘=sin⁡(90∘+10∘)+cos⁡(60∘+10∘)cos⁡(90∘−10∘)−cos⁡(30∘+10∘)=sin⁡90∘cos⁡10∘+cos⁡90∘sin⁡10∘+cos⁡60∘cos⁡10∘−sin⁡60∘sin⁡10∘cos⁡90∘cos⁡10∘+sin⁡90∘sin⁡10∘−cos⁡30∘cos⁡10∘+sin⁡30∘sin⁡10∘=cos⁡10∘+12cos⁡10∘−32sin⁡10∘sin⁡10∘−32cos⁡10∘+12sin⁡10∘=32cos⁡10∘−32sin⁡10∘32sin⁡10∘−32cos⁡10∘

Question

Trigonometry problem sin⁡100∘+cos⁡70∘cos⁡80∘−cos⁡20∘I&#039;ve done trigonometry in my earlier years of high school but I forgot a lot of rules. This is where I&#039;m stuck on this problem:sin⁡100∘+cos⁡70∘cos⁡80∘−cos⁡20∘=sin⁡(90∘+10∘)+cos⁡(60∘+10∘)cos⁡(90∘−10∘)−cos⁡(30∘+10∘)=sin⁡90∘cos⁡10∘+cos⁡90∘sin⁡10∘+cos⁡60∘cos⁡10∘−sin⁡60∘sin⁡10∘cos⁡90∘cos⁡10∘+sin⁡90∘sin⁡10∘−cos⁡30∘cos⁡10∘+sin⁡30∘sin⁡10∘=cos⁡10∘+12cos⁡10∘−32sin⁡10∘sin⁡10∘−32cos⁡10∘+12sin⁡10∘=32cos⁡10∘−32sin⁡10∘32sin⁡10∘−32cos⁡10∘

analiticozuod · Accepted Answer

It is:sin⁡100∘+cos⁡70∘cos⁡80∘−cos⁡20∘=sin⁡80∘+sin⁡20∘cos⁡80∘−cos⁡20∘=2sin⁡50∘cos⁡30∘−2sin⁡50∘sin⁡30∘=−3

seskew192atp · Accepted Answer

Givensin⁡100+cos⁡100cos⁡80−cos⁡20Now to solve the denominator use the formula cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)cos⁡80−cos⁡20=−2sin⁡(80+2020)sin⁡(80−202)=sin⁡100+cos⁡70−2sin⁡(80+2020)sin⁡(80−202)=−sin⁡100+cos⁡70sin⁡50Now to solve the numerator use the identity sin⁡A+sin⁡B=2cos⁡(A−B2)sin⁡(A+B2)sin⁡100+sin⁡20=2cos⁡(100−202)sin⁡(100+202)=−2cos⁡(100−202)sin⁡(100+202)sin⁡50=−2cos⁡40sin⁡60sin⁡50=−2cos⁡40⋅32sin⁡50−3sin⁡(90−40)sin⁡50=−3Therefore,sin⁡100+cos⁡100cos⁡80−cos⁡20=−3

Trigonometry problem \(\displaystyle{\frac{{{{\sin{{100}}}^{{\circ}}+}{\cos{{70}}}^{{\circ}}}}{{{{\cos{{80}}}^{{\circ}}-}{\cos{{20}}}^{{\circ}}}}}\) I've done trigonometry in my

Answered question

Answer & Explanation

New Questions in Precalculus