IF A+B+C=Π prove that sin(3A)+sin(3B)+sin(3C)=-4cos(3A÷2)cos(3B÷2)cos(3C÷2)

Mathematics

1 answer:

Hunter-Best [27]2 years ago

4 0

Answer:

LHS of the given equation is:

sinθ+sin3θ+sin5θ+sin7θ=(sinθ+sin7θ)+(sin3θ+sin5θ)=2sin8θ2.cos6θ2+2sin8θ2.cos2θ2 [since sinC+sinD=2sinC+D2.cosC−D2=2sin4θ.cos3θ+2sin4θ.cosθ=2sin4θ.[cos3θ+cosθ] [since cosC+cosD=2cosC+D2.cosC−D2=2sin4θ.[2cos2θ.cosθ]=4cosθ.cos2θ.sin4θsinθ+sin3θ+sin5θ+sin7θ=(sinθ+sin7θ)+(sin3θ+sin5θ)=2sin8θ2.cos6θ2+2sin8θ2.cos2θ2 [since sinC+sinD=2sinC+D2.cosC-D2=2sin4θ.cos3θ+2sin4θ.cosθ=2sin4θ.[cos3θ+cosθ] [since cosC+cosD=2cosC+D2.cosC-D2=2sin4θ.[2cos2θ.cosθ]=4cosθ.cos2θ.sin4θ

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Step-by-step explanation:

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ABCD is a rhombus, where m∠AED = 5x – 10. Use the properties of a rhombus to determine the value of x. Question 10 options: A) x

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Answer:

A) x = 20

Step-by-step explanation:

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$\therefore \: m \angle AED = 90 \degree \\ \therefore \: (5x - 10) \degree = 90 \degree \\ \therefore \: 5x - 10 = 90 \\ \therefore \: 5x = 90 + 10 \\ \therefore \: 5x = 100 \\ \\ \therefore \: x = \frac{100}{5} \\ \\ \huge \red{ \boxed{ \therefore \: x =20}}$