K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

NV
21 tháng 4 2021

a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)

b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)

c/\(P=sin\left(30+60\right)=sin90=1\)

d/

\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)

\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)

e/

\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)

\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)

21 tháng 4 2021

sina - 1 = sina - sin\(\dfrac{\pi}{2}\)