Sitemap. Trigonometric ratios of angles greater than or equal to 360 degree. \end{array}θsinθcosθtanθ​0∘20​​24​​0​6π​=30∘21​​23​​3​1​​4π​=45∘22​​22​​1​3π​=60∘23​​21​​3​​2π​=90∘24​​20​​∞​​, cos⁡2θ+sin⁡2θ=11+tan⁡2θ=sec⁡2θcot⁡2θ+1=csc⁡2θ.\begin{aligned} \end{aligned}(2​+1)(2​−1)cosθ(2−1)cosθ⇒cosθ−sinθ​=(2​+1)sinθ=2​sinθ+sinθ=2​sinθ. Learn how to solve trigonometric equations and how to use trigonometric identities to solve various problems. \end{aligned} sin(x+y)sin(x−y)cos(x+y)cos(x−y)tan(x+y)tan(x−y)​=sinxcosy+cosxsiny=sinxcosy−cosxsiny=cosxcosy−sinxsiny=cosxcosy+sinxsiny=1−tanxtanytanx+tany​=1+tanxtanytanx−tany​.​, sin⁡3θ=3sin⁡θ−4sin⁡3θcos⁡3θ=4cos⁡3θ−3cos⁡θ.\begin{aligned} Therefore, sin⁡θ=tan⁡θsec⁡θ=−513. & = \sec^2 \theta \cdot \csc^2 \theta.\ _\square \(\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)\), \(\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)\), \(\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\), \(\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\), \(\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\), \(\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}\), \(\sin(x)+\sin(y)=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})\), \(\sin(x)-\sin(y)=2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})\), \(\cos(x)+\cos(y)=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})\), \(\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})\), \(\cos(2x)=\cos^2(x)-\sin^2(x)=1-2\sin^2(x) = 2\cos^2(x)-1\), \(\sin(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{2}}\), \(\cos(\frac{x}{2})=\pm\sqrt{\frac{1+\cos(x)}{2}}\), \(\tan(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\frac{1-\cos(x)}{\sin(x)}=\frac{\sin(x)}{1+\cos(x)}\), \(\sin(x)\cos(y)=\frac{\sin(x+y)+\sin(x-y)}{2}\), \(\cos(x)\cos(y)=\frac{\cos(x+y)+\cos(x-y)}{2}\), \(\sin(x)\sin(y)=\frac{\cos(x-y)-\cos(x+y)}{2}\). Let A  =  tan4θ + tan2θ  and B  =  sec4θ + sec2θ. ... For a problem like sin(π/12), remember that θ/2 = π/12, or θ = π/6, when substituting into the identity. /2, calculate the remaining trigonometric ratios of angle α. & = \tan^2 \theta + \cot^2 \theta + 2\tan \theta \cot \theta \\ Knowing that tan α = 2, and that 180º < α < 270°, calculate the remaining trigonometric ratios of angle α. Note that the three identities above all involve squaring and the number 1.You can see the Pythagorean-Thereom relationship clearly if you consider the unit circle, where the angle is t, the "opposite" side is sin(t) = y, the "adjacent" side is cos(t) = x, and the hypotenuse is 1.. We have additional identities related to the functional status of the trig ratios: □​​. \tan(x-y) &= \dfrac{\tan x - \tan y}{1 + \tan x \tan y}. Lecture Notes Trigonometric Identities 1 page 1 Sample Problems Prove each of the following identities. Knowing that tan α = 2, and that 180º < α <270°, calculate the remaining trigonometric ratios of angle α. By subtracting, (sec⁡θ+tan⁡θ)−(sec⁡θ−tan⁡θ)=23−32  ⟹  2tan⁡θ=−56  ⟹  tan⁡θ=−512(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = \frac23 - \frac32 \implies 2 \tan \theta = \frac{-5}{6} \implies \tan \theta = \frac{-5}{12}(secθ+tanθ)−(secθ−tanθ)=32​−23​⟹2tanθ=6−5​⟹tanθ=12−5​. a^2 & = (4\sin \theta - 3\cos \theta)^2 \\ \\ Knowing that csc α = 3, calculate the remaining trigonometric ratios of angle α. \tan \theta & 0 & \frac { 1}{\sqrt{3} } & 1 & \sqrt{3} & \infty \\ (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. \cot \theta & = \tan \left( \frac{\pi}{2} - \theta \right) \\ & = \dfrac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cdot \cos^2 \theta} \\ \\ & = \sec^2 \theta + \csc^2 \theta Khan Academy is a 501(c)(3) nonprofit organization. Let A  =  (1 - sin A)/(1 + sin A) and B  =  (sec A - tan A)2. 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∘​. \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} Sign up, Existing user? Let A  =  sec θ √(1 - sin2θ)  and B  =  1. Knowing that cos α = ¼ , and that 270º <α <360°, calculate the remaining trigonometric ratios of angle α. &=\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) \\ & = \tan^2 \theta + \cot^2 \theta + 2 \\ □\begin{aligned} 2(sin⁡6θ+cos⁡6θ)−3(sin⁡4θ+cos⁡4θ)=2[(sin⁡2θ)3+(cos⁡2θ)3]−3[(sin⁡2θ)2+(cos⁡2θ)2]=2[(sin⁡2θ+cos⁡2θ)3−3sin⁡2θcos⁡2θ(sin⁡2θ+cos⁡2θ)]−3[(sin⁡2θ+cos⁡2θ)2−sin⁡2θcos⁡2θ]=2(1−3sin⁡2θcos⁡2θ)−3(1−2sin⁡2θcos⁡2θ)=2−3=−1. \hline Calculate the trigonometric ratios of 15 (from the 45º and 30º). sin⁡2θ+cos⁡2θ=1cos⁡2θ=1−sin⁡2θ=1−1625=925⇒cos⁡θ=±35⇒cos⁡θ=−35. A  =   (1 - sin A)2 / (1 - sin A) (1 + sin A), (tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ, Let A  =  (tan θ + sec θ - 1)/(tan θ - sec θ + 1)  and, A  =  (tan θ + sec θ - 1)/(tan θ - sec θ + 1), A  =  [(tan θ + sec θ) - (sec2θ - tan2θ)]/(tan θ - sec θ + 1), A  =  {(tan θ + sec θ) (1 - sec θ + tan θ)}/(tan θ - sec θ + 1), A  =  {(tan θ + sec θ) (tan θ - sec θ + 1)}/(tan θ - sec θ + 1). A  =  âˆš[{(sec θ - 1) (sec θ - 1)}/{(sec θ + 1) (sec θ - 1)}], A  =  {(1/cos θ)/(sin θ/cos θ)} - cot θ, A  =  {(1/cos θ) â‹… (cos θ/sin θ)} - cot θ, (1 - sin A)/(1 + sin A)  =  (sec A - tan A)2. Try the free Mathway calculator and \cot \dfrac{\pi}{16} \cdot \cot \dfrac{2 \pi}{16} \cdot \cot \dfrac{3 \pi}{16} \times \cdots \times \cot \dfrac{7 \pi}{16} \end{aligned}sinx+sinycosx+cosy​=2sin(2x+y​)cos(2x−y​)=2cos(2x+y​)cos(2x−y​).​. \cos \theta & = \sin \left( \frac{\pi}{2} - \theta \right) \\ The residents home is perpendicular to the ground. Sign up to read all wikis and quizzes in math, science, and engineering topics. \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ \end{aligned} sin3θcos3θ​=3sinθ−4sin3θ=4cos3θ−3cosθ.​. \end{aligned}52a252+a2a2+25a2+25a2+25a⇒4sinθ−3cosθ​=(3sinθ+4cosθ)2=(4sinθ−3cosθ)2=(3sinθ+4cosθ)2+(4sinθ−3cosθ)2=9sin2θ+16cos2θ+24sinθcosθ+16sin2θ+9cos2θ−24sinθcosθ=25(sin2θ+cos2θ)=25=0=0. I am passionate about travelling and currently live and work in Paris. &=\underbrace{\left(\cot \dfrac{\pi}{16} \cdot \tan \dfrac{\pi}{16}\right)}_{1} \cdot \underbrace{\left(\cot \dfrac{2 \pi}{16} \cdot \tan \dfrac{2\pi}{16} \right)}_{1} \cdot \underbrace{\left(\cot \dfrac{3 \pi}{16} \cdot \tan \dfrac{3\pi}{16} \right)}_{1} \cdot 1 \\ Quotient Identities. \sin(x-y) &= \sin x \cos y - \cos x \sin y \\\\ &= 1.\ _\square Check out all of our online calculators here! sin⁡xsin⁡y=12,cos⁡xcos⁡y=32, \frac {\sin x}{\sin y } = \frac {1}{2}, \quad \frac {\cos x}{\cos y } = \frac 3 2 , sinysinx​=21​,cosycosx​=23​. Our mission is to provide a free, world-class education to anyone, anywhere. \hline \sin x \cos y &= \frac{1}{2} \big(\sin (x- y) + \sin(x + y) \big) \\ √{(sec θ – 1)/(sec θ + 1)}  =  cosec θ - cot θ. Knowing that sec α = 2 and 0 < α < what is tan⁡2(x+y)? Find the value of 2(sin⁡6θ+cos⁡6θ)−3(sin⁡4θ+cos⁡4θ)2\big(\sin^6 \theta + \cos^6 \theta\big) - 3\big(\sin^4 \theta + \cos^4 \theta\big)2(sin6θ+cos6θ)−3(sin4θ+cos4θ). &=\underbrace{\sin^2 \dfrac{\pi}{10} + \cos^2 \dfrac{\pi}{10}}_{1} + \underbrace{\cos^2 \dfrac{\pi}{10} + \sin^2 \dfrac{\pi}{10}}_{1} \\ Find the value of cot⁡π16×cot⁡2π16×cot⁡3π16×⋯×cot⁡7π16\cot \frac{\pi}{16} \times \cot \frac{2 \pi}{16} \times \cot \frac{3 \pi}{16} \times \cdots \times \cot \frac{7 \pi}{16}cot16π​×cot162π​×cot163π​×⋯×cot167π​. Use these fundemental formulas of trigonometry to help solve problems by re … □_\square□​. \cos x+\cos y &=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right). Already have an account? Domain and range of trigonometric … \end{aligned} sin2A+cos2Atan2A+1cot2A+1​===​1sec2Acsc2A.​, sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B.\begin{aligned} \sin(A\pm B) &=& \sin A \cos B \pm \cos A \sin B \\ \cos(A\pm B) &=& \cos A \cos B \mp \sin A \sin B \\ \tan(A\pm B) &=& \frac{ \tan A \pm \tan B } { 1 \mp \tan A \tan B }. (2+1)(2−1)cos⁡θ=(2+1)sin⁡θ(2−1)cos⁡θ=2sin⁡θ+sin⁡θ⇒cos⁡θ−sin⁡θ=2sin⁡θ. & = \dfrac{1}{\sin^2 \theta \cdot \cos^2 \theta} \\ \\ □​​. If sin⁡θ=45\sin \theta = \frac45sinθ=54​ and θ\thetaθ is not in the first quadrant, then find the value of cos⁡θ\cos \thetacosθ. \sin 3 \theta &= 3 \sin \theta - 4 \sin ^3 \theta \\ □​​. A  =  (cos2θ/sin θ cos Î¸) + (sin2θ/sin Î¸ cos Î¸), cos θ/(1 - tan θ) + sin θ/(1 - cot θ)  =  sin θ + cos θ, Let A  =  cos θ/(1 - tan θ) + sin θ/(1 - cot θ)  and, A  =  cos θ/{1 - (sin θ/cos θ)} + sin θ/{1 - (cos θ/sin θ)}, A  =  cos2θ/(cos θ - sin θ) + sin2θ/(sin θ - cos θ), A  =  cos2θ/(cos θ - sin θ) - sin2θ/(cos θ - sin θ), A  =  (cos2θ - sin2θ) / (cos θ - sin θ), A  =  [(cos θ + sin θ)(cos θ - sin θ)] / (cos θ - sin θ). &=\left(\cot \dfrac{\pi}{16} \cdot \cot \dfrac{7 \pi}{16}\right) \cdot \left(\cot \dfrac{2 \pi}{16} \cdot \cot \dfrac{6 \pi}{16} \right) \cdot \left(\cot \dfrac{3 \pi}{16} \cdot \cot \dfrac{5 \pi}{16} \right) \cdot \cot \dfrac{4 \pi}{16} \\ To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Quotient Identities. &= 1 - 2\sin^2 \theta \\\\ Trigonometric ratios of complementary angles. \cos \theta &=\cos(\theta+2\pi) &\quad \sec \theta &=\sec(\theta+2\pi)\\ \sin^2 \theta &=\frac{1}{2}(1-\cos 2\theta) \\ \tan^2 (x+y)?tan2(x+y)? Problem : What is sin(- )? Let A  =  (1 - cos2θ) csc2θ  and  B  =  1. If a,b,a, b,a,b, and ccc are constants that satisfy the trigonometric identity above, find the value of c.c.c. & = 2 - 3 \\ \sin 2 \theta &= 2\sin \theta \cos \theta \\\\ □​​, 1. Identities In Use‎ > ‎ Word Problem During a stormy night in Louisiana, a tree fell on a residents home. □​​. Calculate the trigonometric ratios of 15º (from the 45º and 30º). Our mission is to provide a free, world-class education to anyone, anywhere. Let A  =  tan Î¸ sin θ + cos θ  and B =  sec θ. &= 2\cos^2 \theta - 1\\ Log in. □(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​. Let A  =  (1 - cos θ)(1 + cos θ)(1 + cot2θ)  =  1 and B  =  1.