\sin(x-y) &= \sin x \cos y - \cos x \sin y \\\\ Already have an account? □​​. Trigonometric identities (trig identities) are equalities that involve trigonometric functions that are true for all values of the occurring variables. Log in here. \sin \theta &= \cos \left( \frac{\pi}{2}-\theta \right) \\

Sitemap. sin⁡2A+cos⁡2A=1tan⁡2A+1=sec⁡2Acot⁡2A+1=csc⁡2A.\begin{aligned} \sin^2 A + \cos^2 A &=& 1 \\ \tan^2 A + 1 &=& \sec^2 A \\ \cot^2 A + 1 &=& \csc^2 A. Draw a picture illustrating the problem if it involves only the basic trigonometric functions.

\tan(x-y) &= \dfrac{\tan x - \tan y}{1 + \tan x \tan y}. Knowing that csc α = 3, calculate the remaining trigonometric ratios of angle α. \end{aligned}cos(−θ)sin(−θ)tan(−θ)cot(−θ)csc(−θ)sec(−θ)​=cosθ=−sinθ=−tanθ=−cotθ=−cscθ=secθ.​, sin⁡θ=sin⁡(θ+2π)csc⁡θ=csc⁡(θ+2π)cos⁡θ=cos⁡(θ+2π)sec⁡θ=sec⁡(θ+2π)tan⁡θ=tan⁡(θ+π)cot⁡θ=cot⁡(θ+π).\begin{aligned} & = -1.\ _\square

/2, calculate the remaining trigonometric ratios of angle α. Embedded content, if any, are copyrights of their respective owners. \cot^2 \theta + 1 &= \csc^2 \theta. □\begin{aligned} &=\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} \\ \cos(-\theta) &= \cos \theta \\ Knowing that sec α = 2 and 0< α < Problem : What is sin(- )? & = \dfrac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cdot \cos^2 \theta} \\ \\ \cos(x+y) &= \cos x \cos y - \sin x \sin y \\ \sin x \cos y &= \frac{1}{2} \big(\sin (x- y) + \sin(x + y) \big) \\ Trigonometric ratios of supplementary angles Trigonometric identities Problems on trigonometric identities Trigonometry heights and distances. Reciprocal Identities. A  =  (1 - cos θ)(1 + cos θ)(1 + cot2θ), A  =  sin2θ  + sin2θ â‹… (cos2θ/sin2θ).

2.\ \sec^2 \theta + \csc^2 \theta \cos x+\cos y &=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right). The tree is 42.5 feet tall and the base of the tree is 28 feet away from the house.

Reciprocal Identities. 1. tanxsinx+cosx = secx 2.

□\begin{aligned}

\sin(x+y) &= \sin x \cos y + \cos x \sin y \\

Therefore, sin⁡θ=tan⁡θsec⁡θ=−513.

Calculate the trigonometric ratios of 15 (from the 45º and 30º). The residents home is perpendicular to the ground. \cos 3\theta &= 4 \cos ^ 3 \theta - 3 \cos \theta. & = \big(1 + \tan^2 \theta\big) + \big(1 + \cot^2 \theta\big) \\ \cos \theta & \frac {\sqrt{4}} {2} & \frac {\sqrt{3}} {2} & \frac {\sqrt{2}} {2} & \frac {\sqrt{1}} {2} & \frac {\sqrt{0}} {2}\\ □\begin{aligned} If cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta + \sin \theta = \sqrt{2} \cos \thetacosθ+sinθ=2​cosθ, find the value of cos⁡θ−sin⁡θ\cos \theta - \sin \thetacosθ−sinθ. & = 2\left[\big(\sin^2 \theta + \cos^2 \theta\big)^3 - 3\sin^2 \theta \cos^2 \theta\big(\sin^2 \theta + \cos^2 \theta\big)\right] - 3\left[\big(\sin^2 \theta + \cos^2 \theta\big)^2 - \sin^2 \theta \cos^2 \theta\right] \\ /2, calculate the remaining trigonometric ratios of angle α. Knowing that cos α = ¼ , and that 270º <α <360°, calculate the remaining trigonometric ratios of angle α. & = 2\big(1 - 3\sin^2 \theta \cos^2 \theta\big) - 3\big(1 - 2\sin^2 \theta \cos^2 \theta\big) \\ \sin^2 \dfrac{\pi}{10} + \sin^2 \dfrac{4\pi}{10} + \sin^2 \dfrac{6 \pi}{10} + \sin^2 \dfrac{9 \pi}{10}

\cot \dfrac{\pi}{16} \cdot \cot \dfrac{2 \pi}{16} \cdot \cot \dfrac{3 \pi}{16} \times \cdots \times \cot \dfrac{7 \pi}{16} \cos \theta &=\cos(\theta+2\pi) &\quad \sec \theta &=\sec(\theta+2\pi)\\ I like to spend my time reading, gardening, running, learning languages and exploring new places. □​​. 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). Our mission is to provide a free, world-class education to anyone, anywhere. Sign up, Existing user? Let A  =  âˆš{(sec θ – 1)/(sec θ + 1)} and B  =  cosec θ - cot θ.

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. Sign up to read all wikis and quizzes in math, science, and engineering topics. &= 1.\ _\square □\begin{aligned} 5^2 + a^2 & = (3\sin \theta + 4 \cos \theta)^2 + (4\sin \theta - 3\cos \theta)^2 \\ \tan 2\theta &= \frac{2\tan \theta}{1 - \tan^2 \theta}. Log in.

\tan(-\theta) &= -\tan \theta\\ 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∘​. & = \sec^2 \theta \cdot \csc^2 \theta.\ _\square

Let A  =  (1 - sin A)/(1 + sin A) and B  =  (sec A - tan A)2. \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} New user? The following is a list of useful Trigonometric identities: Quotient Identities, Reciprocal Identities, Pythagorean Identities, Co-function Identities, Addition Formulas, Subtraction Formulas, Double Angle Formulas, Even Odd Identities, Sum-to-product formulas, Product-to-sum formulas. Knowing that tan α = 2, and that 180º < α <270°, calculate the remaining trigonometric ratios of angle α.

&= 2\cos^2 \theta - 1\\ Use these fundemental formulas of trigonometry to help solve problems by re … \theta & 0^\circ & \frac{\pi}{6} = 30^\circ & \frac{\pi}{4} = 45^\circ & \frac{\pi}{3} = 60^\circ & \frac{\pi}{2} = 90^\circ\\ Problems; Additional Trigonometric Identities; Problems; Review of Functions and Angles; Problems; Key Terms; Writing Help. A comprehensive list of the important trigonometric identity formulas.

& = \tan^2 \theta + \cot^2 \theta + 2 \\