Find the value of the expression sin⁡(135∘−2α)⋅sin⁡(α−45∘)+cos2(75∘−α)−2sin⁡(α−30∘) when α=60∘ when α=60∘ I don't know if I am supposed to simplify the given expression, but I just put α=60∘ to get sin⁡(135∘−120∘)⋅sin⁡(60∘+45∘)+cos2(75∘−60∘)−2sin⁡(60∘−30∘)= =sin⁡15∘sin⁡105∘+cos⁡215∘−2sin⁡30∘ What can I do next?

Question

Find the value of the expression sin⁡(135∘−2α)⋅sin⁡(α−45∘)+cos2(75∘−α)−2sin⁡(α−30∘) when α=60∘ when α=60∘ I don&#039;t know if I am supposed to simplify the given expression, but I just put α=60∘ to get sin⁡(135∘−120∘)⋅sin⁡(60∘+45∘)+cos2(75∘−60∘)−2sin⁡(60∘−30∘)=  =sin⁡15∘sin⁡105∘+cos⁡215∘−2sin⁡30∘  What can I do next?

Mary Goodson · Accepted Answer

This seems to be a simpler question than what is thought to be.   Replace α with 60∘, just remember   sin⁡30∘=12,  sin2x+cos2x=1   Then,  sin⁡15∘sin⁡15∘+cos215∘−2sin⁡30∘  =1−2⋅12=0

Suhadolahbb · Accepted Answer

HINT   You could first use the following formula, known as one of the product to sum formulae among other names,   sin⁡xsin⁡y≡12cos⁡(x−y)−cos⁡(x+y)  to evaluate sin⁡15∘sin⁡105∘ without having to explicitly evaluate sin⁡15∘ and sin⁡105∘  You could then use the double angle formula for cos⁡ to write cos⁡215∘ in terms of  cos⁡30∘

Vasquez · Accepted Answer

Hint: Notice that sin⁡105∘=cos⁡15∘. Then you have to use the formulae: cos⁡2θ+12=cos2⁡θ And, sin⁡2θ=2sin⁡θcos⁡θ If you do it this way, you won&#039;t have to calculate trigonometric ratios for 15∘,which can be evaluated, but are slightly complicated.

Find the value of the expression \sin(135^{\circ}-2\alpha) \cdot \sin(\alpha-45^{\circ})+\cos^2(75^{\circ}-\alpha)-2\sin(\alpha-30^{\circ}) \

Answered question

Answer & Explanation

New Questions in Other