Tính giá trị của biểu thức P= sin(-14π/3) + 1/(sin²29π/4) – tan²(3π/4) 28/09/2021 Bởi Eden Tính giá trị của biểu thức P= sin(-14π/3) + 1/(sin²29π/4) – tan²(3π/4)
Đáp án: $ P=-\dfrac{\sqrt{3}}{2}+3$ Giải thích các bước giải: Ta có : $\sin\left(\dfrac{-14\pi}{3}\right)=-\sin \left(\dfrac{14\pi }{3}\right)=-\sin \left(\dfrac{2\pi }{3}+4\pi\right)=-\sin \left(\dfrac{2\pi }{3}\right)=-\dfrac{\sqrt{3}}{2}$ $\sin\left(\dfrac{29\pi}{4}\right)=\sin \left(\dfrac{5\pi }{4}+6\pi\right)=\sin \left(\dfrac{5\pi }{4}\right)=\sin \left(\dfrac{\pi }{4}+\pi\right)=-\sin \left(\dfrac{\pi }{4}\right)=-\dfrac{\sqrt{2}}{2}$ $\tan\left(\dfrac{3\pi}{4}\right)=\tan\left(\pi-\dfrac{\pi}{4}\right)=\tan\left(-\dfrac{\pi}{4}\right)=-\tan\left(\dfrac{\pi}{4}\right)=-1$ $\to P=-\dfrac{\sqrt{3}}{2}+3$ Bình luận
$P=\dfrac{\sin(\frac{-14\pi}{3})+1}{\sin^2\dfrac{29\pi}{4}-\tan^2\dfrac{3\pi}{4}}$ Ta có: $\sin\dfrac{-14\pi}{3}=\sin(-4\pi-\dfrac{2\pi}{3}=\sin(-\dfrac{2\pi}{3})=\dfrac{-\sqrt3}{2}$ $\sin^2\dfrac{29\pi}{4}=\sin^2(6\pi+\pi+\dfrac{\pi}{4}=\sin^2\dfrac{\pi}{4}=\dfrac{1}{2}$ $\tan^2\dfrac{3\pi}{4}=\tan^2\dfrac{\pi}{4}=1$ $\Rightarrow P=\dfrac{2-\sqrt3}{3}$ Bình luận
Đáp án: $ P=-\dfrac{\sqrt{3}}{2}+3$
Giải thích các bước giải:
Ta có :
$\sin\left(\dfrac{-14\pi}{3}\right)=-\sin \left(\dfrac{14\pi }{3}\right)=-\sin \left(\dfrac{2\pi }{3}+4\pi\right)=-\sin \left(\dfrac{2\pi }{3}\right)=-\dfrac{\sqrt{3}}{2}$
$\sin\left(\dfrac{29\pi}{4}\right)=\sin \left(\dfrac{5\pi }{4}+6\pi\right)=\sin \left(\dfrac{5\pi }{4}\right)=\sin \left(\dfrac{\pi }{4}+\pi\right)=-\sin \left(\dfrac{\pi }{4}\right)=-\dfrac{\sqrt{2}}{2}$
$\tan\left(\dfrac{3\pi}{4}\right)=\tan\left(\pi-\dfrac{\pi}{4}\right)=\tan\left(-\dfrac{\pi}{4}\right)=-\tan\left(\dfrac{\pi}{4}\right)=-1$
$\to P=-\dfrac{\sqrt{3}}{2}+3$
$P=\dfrac{\sin(\frac{-14\pi}{3})+1}{\sin^2\dfrac{29\pi}{4}-\tan^2\dfrac{3\pi}{4}}$
Ta có:
$\sin\dfrac{-14\pi}{3}=\sin(-4\pi-\dfrac{2\pi}{3}=\sin(-\dfrac{2\pi}{3})=\dfrac{-\sqrt3}{2}$
$\sin^2\dfrac{29\pi}{4}=\sin^2(6\pi+\pi+\dfrac{\pi}{4}=\sin^2\dfrac{\pi}{4}=\dfrac{1}{2}$
$\tan^2\dfrac{3\pi}{4}=\tan^2\dfrac{\pi}{4}=1$
$\Rightarrow P=\dfrac{2-\sqrt3}{3}$