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Vì 300 và 600 là hai góc phụ nhau nên sin 30 0 = cos 60 0 sin 60 0 = cos 30 0
⇒ P = cos 30 ∘ cos 60 ∘ − sin 30 ∘ sin 60 ∘ = cos 30 ∘ cos 60 ∘ − cos 60 ∘ cos 30 ∘ = 0.
Chọn D.
Vì 300 và 600 là hai góc phụ nhau nên sin 30 0 = cos 60 0 sin 60 0 = cos 30 0
⇒ P = sin 30 ∘ cos 60 ∘ + sin 60 ∘ cos 30 ∘ = cos 2 60 ∘ + sin 2 60 ∘ = 1.
Chọn A.
Vì 300 và 600 là hai góc phụ nhau nên sin 30 0 = cos 60 0 sin 60 0 = cos 30 0
⇒ P = cos 30 ∘ cos 60 ∘ − sin 30 ∘ sin 60 ∘ = cos 30 ∘ cos 60 ∘ − cos 60 ∘ cos 30 ∘ = 0.
Chọn D.
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/
\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)
\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)
a)
\(A=cos^230^o-sin^230^o=\left(\dfrac{\sqrt{3}}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2}\);
\(B=cos60^o+sin45^o=\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}\).
Vì vậy \(A< B\).
b)
\(C=\dfrac{2tan30^o}{1-tan^230^o}=\dfrac{2\dfrac{\sqrt{3}}{2}}{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}=\sqrt{3}\).
\(D=\left(-tan135^o\right)tan60^o=-\left(-1\right).\sqrt{3}=\sqrt{3}\).
Vậy \(C=D\).
\(A=cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\left(-cos\left(\pi-\dfrac{5\pi}{7}\right)\right)=-cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(\Rightarrow A.sin\left(\dfrac{\pi}{7}\right)=-sin\left(\dfrac{\pi}{7}\right).cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=-\dfrac{1}{2}sin\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)=-\dfrac{1}{4}sin\left(\dfrac{4\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)
\(=-\dfrac{1}{8}sin\left(\dfrac{8\pi}{7}\right)=\dfrac{1}{8}sin\left(\dfrac{\pi}{7}\right)\)
\(\Rightarrow A=\dfrac{1}{8}\)
\(B=\dfrac{\sqrt{3}}{2}.cos48^0.cos24^0.cos12^0\)
\(\Rightarrow B.sin12^0=\dfrac{\sqrt{3}}{2}sin12^0.cos12^0cos24^0.cos48^0\)
\(=\dfrac{\sqrt{3}}{4}sin24^0cos24^0cos48^0=\dfrac{\sqrt{3}}{8}sin48^0.cos48^0\)
\(=\dfrac{\sqrt{3}}{16}sin96^0=\dfrac{\sqrt{3}}{16}cos6^0\)
\(\Rightarrow2B.sin6^0.cos6^0=\dfrac{\sqrt{3}}{16}cos6^0\Rightarrow B=\dfrac{\sqrt{3}}{32.sin6^0}\)
Biểu thức này ko thể rút gọn tiếp được
Chú ý rằng: sin450 = cos450, sin400 = cos500, sin500 = cos400
Ta được:
\(\dfrac{\cos50^0-\cos45^0+\cos50^0}{\cos40^0-\cos45^0+\cos50^0}-\dfrac{6\times3\left(\dfrac{\sqrt{3}}{3}+\tan15^0\right)}{3\left(1-\dfrac{\sqrt{3}}{3}\tan15^0\right)}\)
\(=1-6\left(\dfrac{\tan30^0+\tan15^0}{1-\tan30^0\times\tan15^0}\right)\)
\(=1-6\tan45^0=-5\)
a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).
\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).
Ta có sin0 độ + cos0 độ =0+1=1 nên A sai.
sin90 độ + cos90 độ =1+0=1 nên B đúng.
sin180 độ + cos180 độ=0+(−1)=−1 nên C đúng.
sin60 độ + cos60 độ =√3/2+1/2=√3+1/2 nên D đúng.
Chọn (A).
a)
\(2sin30+3sin45^o-sin60^o=2.\dfrac{1}{2}+3.\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}\)\(=\dfrac{2+3\sqrt{2}-\sqrt{3}}{2}\).
b)\(2cos30^o+3sin45^o-cos60^o=2.\dfrac{\sqrt{3}}{2}+3.\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\)\(=\dfrac{2\sqrt{3}+3\sqrt{2}-1}{2}\).