\end{aligned}cosθsinθcotθcscθ​=sin(2π​−θ)=cos(2π​−θ)=tan(2π​−θ)=sec(2π​−θ).​, sin⁡2θ=2sin⁡θcos⁡θcos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θtan⁡2θ=2tan⁡θ1−tan⁡2θ.\begin{aligned} Multiply rational expressions by conjugates in order to take advantage of the. \end{aligned} sin2θcos2θ​=21​(1−cos2θ)=21​(1+cos2θ).​, cos⁡xcos⁡y=12(cos⁡(x−y)+cos⁡(x+y))sin⁡xcos⁡y=12(sin⁡(x−y)+sin⁡(x+y))cos⁡xsin⁡y=12(sin⁡(x+y)−sin⁡(x−y))sin⁡xsin⁡y=12(cos⁡(x−y)−cos⁡(x+y)).\begin{aligned} \tan(x-y) &= \dfrac{\tan x - \tan y}{1 + \tan x \tan y}. sin⁡θ=2tan⁡θ21+tan⁡2θ2cos⁡θ=1−tan⁡2θ21+tan⁡2θ2tan⁡θ=2tan⁡θ21−tan⁡2θ2.\sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^{2}\frac{\theta}{2}} \\ \cos\theta=\frac{1-\tan^{2}\frac{\theta}{2}}{1+\tan^{2}\frac{\theta}{2}} \\ \tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^{2}\frac{\theta}{2}}.sinθ=1+tan22θ​2tan2θ​​cosθ=1+tan22θ​1−tan22θ​​tanθ=1−tan22θ​2tan2θ​​. \end{aligned}cosxcosysinxcosycosxsinysinxsiny​=21​(cos(x−y)+cos(x+y))=21​(sin(x−y)+sin(x+y))=21​(sin(x+y)−sin(x−y))=21​(cos(x−y)−cos(x+y)).​, sin⁡x+sin⁡y=2sin⁡(x+y2)cos⁡(x−y2)cos⁡x+cos⁡y=2cos⁡(x+y2)cos⁡(x−y2).\begin{aligned}

&=\sin^2 \left(\dfrac{\pi}{10}\right) + \sin^2 \left( \dfrac{\pi}{2} - \dfrac{\pi}{10} \right) + \sin^2 \left(\dfrac{\pi}{2} + \dfrac{\pi}{10} \right) + \sin^2 \left(\pi - \dfrac{\pi}{10} \right) \\ New user? Let A  =  (1 - cos2θ) csc2θ  and  B  =  1. cos⁡475∘+sin⁡475∘+3sin⁡275∘cos⁡275∘cos⁡675∘+sin⁡675∘+4sin⁡275∘cos⁡275∘.\frac{\cos^4 75^{\circ}+\sin^4 75^{\circ}+3\sin^2 75^{\circ}\cos^2 75^{\circ}}{\cos^6 75^{\circ}+\sin^6 75^{\circ}+4\sin^2 75^{\circ}\cos^2 75^{\circ}}.cos675∘+sin675∘+4sin275∘cos275∘cos475∘+sin475∘+3sin275∘cos275∘​. Therefore, sin⁡θ=tan⁡θsec⁡θ=−513. Reciprocal Identities.

\sin x \sin y &= \frac{1}{2} \big(\cos (x - y) - \cos(x + y) \big).

\sin(-\theta) &= -\sin \theta\\

sin⁡2π10+sin⁡24π10+sin⁡26π10+sin⁡29π10=sin⁡2(π10)+sin⁡2(π2−π10)+sin⁡2(π2+π10)+sin⁡2(π−π10)=sin⁡2π10+cos⁡2π10⏟1+cos⁡2π10+sin⁡2π10⏟1=1+1=2. a^2 + 25 & = 25(\sin^2 \theta + \cos^2 \theta) \\ \end{aligned}2(sin6θ+cos6θ)−3(sin4θ+cos4θ)​=2[(sin2θ)3+(cos2θ)3]−3[(sin2θ)2+(cos2θ)2]=2[(sin2θ+cos2θ)3−3sin2θcos2θ(sin2θ+cos2θ)]−3[(sin2θ+cos2θ)2−sin2θcos2θ]=2(1−3sin2θcos2θ)−3(1−2sin2θcos2θ)=2−3=−1. 1 + \tan^2 \theta &= \sec^2 \theta \\

Reciprocal Identities. Lecture Notes Trigonometric Identities 1 page 1 Sample Problems Prove each of the following identities. 2.\ \sec^2 \theta + \csc^2 \theta (tanθ+cotθ)2​=tan2θ+cot2θ+2tanθcotθ=tan2θ+cot2θ+2=(1+tan2θ)+(1+cot2θ)=sec2θ+csc2θ​, 2. sec⁡2θ+csc⁡2θ=1cos⁡2θ+1sin⁡2θ=sin⁡2θ+cos⁡2θsin⁡2θ⋅cos⁡2θ=1sin⁡2θ⋅cos⁡2θ=sec⁡2θ⋅csc⁡2θ. &=1 + 1 \\

\end{aligned}sin210π​+sin2104π​+sin2106π​+sin2109π​​=sin2(10π​)+sin2(2π​−10π​)+sin2(2π​+10π​)+sin2(π−10π​)=1sin210π​+cos210π​​​+1cos210π​+sin210π​​​=1+1=2.

\end{aligned} sin2θcos2θtan2θ​=2sinθcosθ=cos2θ−sin2θ=2cos2θ−1=1−2sin2θ=1−tan2θ2tanθ​.​, sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡ysin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡ycos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡ycos⁡(x−y)=cos⁡xcos⁡y+sin⁡xsin⁡ytan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡ytan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y.\begin{aligned} If a,b,a, b,a,b, and ccc are constants that satisfy the trigonometric identity above, find the value of c.c.c.

Log in here. Trigonometric ratios of supplementary angles Trigonometric identities Problems on trigonometric identities Trigonometry heights and distances Knowing that sec α = 2 and 0< α < Calculate the trigonometric ratios of 15 (from the 45º and 30º). Trigonometric ratios of complementary angles. The derivations of the half-angle identities … □(since cos⁡θ<0)\begin{aligned} \sin^2 \theta + \cos^2 \theta & = 1 \\ \cos^2 \theta & = 1 - \sin^2 \theta \\ & = 1 - \dfrac{16}{25} = \dfrac{9}{25} \\ \Rightarrow \cos \theta & = \pm \dfrac35 \\ \Rightarrow \cos \theta & = - \dfrac35.\ _\square \qquad (\text{since } \cos \theta < 0) \\ \end{aligned}sin2θ+cos2θcos2θ⇒cosθ⇒cosθ​=1=1−sin2θ=1−2516​=259​=±53​=−53​. a^2 + 25 & = 9\sin^2 \theta + 16\cos^2 \theta + 24\sin\theta\cos\theta + 16\sin^2\theta + 9\cos^2\theta - 24\sin\theta\cos\theta \\

Knowing that cos α = ¼ , and that 270º < α < 360°, calculate the remaining trigonometric ratios of angle α. Quotient Identities. \cos^2 \theta + \sin^2 \theta &= 1 \\ Let A  =  (1 - cos θ)(1 + cos θ)(1 + cot2θ)  =  1 and B  =  1. Problem : What is sin(- )? Trigonometric Identities. &= 2.\ _\square

Let A  =  âˆš{(sec θ – 1)/(sec θ + 1)} and B  =  cosec θ - cot θ.

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Trigonometric ratios of 270 degree minus theta. Please submit your feedback or enquiries via our Feedback page. \end{aligned}1. \sin(x+y) &= \sin x \cos y + \cos x \sin y \\

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