If 2sin theta = 2- cos theta, what is sin theta?
[math]2\sin \theta = 2 - \cos \theta \Rightarrow \cos \theta = 2 - 2\sin \theta \Rightarrow {\cos ^2}\theta = 4 - 8\sin \theta + 4{\sin ^2}\theta \Rightarrow 5{\sin ^2}\theta - 8\sin \theta + 3 = 0 \Rightarrow \sin \theta = 0.6,\quad 1[/math]
If 7 sin square theta + 3 cos square theta is equal to 4. How do you prove that tan theta is equal to 1 by root 3?
Let us assume theta=x for convenience.7 sin^2 x + 3 cos^ 2 x = 4 … eqn 1 - given4 sin ^2 x = 1sin^2 x = 1/4. …… eqn 27 - 4 cos^ 2 x = 4.cos^2 x = 3/4. eqn 3.Divide eqn 2 by eqn 3.tan^2 x = (1/4)/(3/4)Square rooting & putting back x=theta.Tan theta = sqrt (1/3).Sanjay C.
Find all solutions of sin theta+cos theta=-square root 2?
sin^2 theta + 2 sin theta cos theta + cos^2 theta = 2 1 + sin 2theta = 2 sin 2theta = 1 2theta = ( pi / 2 ) + 2n*pi theta = ( pi / 4 ) + n*pi
If sin theta - cos theta = 3/5,what does sin theta × cos theta equal?
(Sin theta-Cos theta)²==sin² theta+cos²theta-2sin theta*Cos theta9/25=1–2sin theta* cos theta2sin theta *cos theta=1–9/25=16/25sin theta*cos theta=16/50
Sin (theta) * Cos (theta), Solve for theta?
I'm having trouble solving for theta in the following equation. I thought there would be some trigonometric identity, but It is lost on me. Any help would be appreciated. 0.964404559=Sin (theta) * Cos (theta) Solve for possible values of Theta
Can we find [math]\theta[/math] if [math]\sin(\theta) + \cos(\theta) = x[/math]?
here it is
If [math]\sin\theta+\cos\theta=\sqrt 2[/math] , what is [math]\tan\theta+\cot\theta[/math] ?
[math]\qquad\:\:\sin\theta+\cos\theta=\sqrt{2}[/math][math]\implies\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=2[/math][math]\implies 1+2\sin\theta\cos\theta=2[/math][math]\implies\sin\theta\cos\theta=\frac{1}{2}[/math][math]\quad\tan\theta+\cot\theta[/math][math]=\displaystyle\frac{\sin\theta}{\cos\theta}+\displaystyle\frac{\cos\theta}{\sin\theta}[/math][math]=\displaystyle\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}[/math][math]=\displaystyle\frac{1}{\frac{1}{2}}[/math][math]=2[/math]
How do I solve cos theta + sin theta = root 2?
Given, [math]\cos\theta + \sin\theta = \sqrt{2}[/math]Also, [math]\cos\theta = \sqrt{2} - \sin\theta[/math] ……. (i)Now, we solve [math]\cos\theta + \sin\theta = \sqrt{2},[/math]Squaring both sides,[math](\cos\theta + \sin\theta)^2 = (\sqrt{2})^2[/math][math]\cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta = 2[/math][math]1 + 2\cos\theta\sin\theta = 2[/math] [Since, [math]\sin^2\theta + \cos^2\theta = 1[/math]][math]1 + 2\cdot(\sqrt{2} - \sin\theta)\cdot\sin\theta = 2[/math] [From (i)][math]1 + 2\sqrt{2}\sin\theta - 2\sin^2\theta = 2[/math]Re-arranging the equation,[math]2\sin^2\theta - 2\sqrt{2}\sin\theta + 1 = 0[/math]([math]\sqrt{2}\sin\theta)^2 - 2\cdot\sqrt{2}\sin\theta\cdot1 + (1)^2 = 0[/math][math](\sqrt{2}\sin\theta - 1)^2 = 0[/math]Square rooting both sides,[math]\sqrt{2}\sin\theta - 1 = 0[/math][math]\sqrt{2}\sin\theta = 1[/math][math]\sin\theta = \frac{1}{\sqrt{2}}[/math][math]\sin\theta = \sin45°[/math] [Since, [math]\sin45° = \frac{1}{\sqrt{2}}][/math]Eliminating [math]\sin[/math] from both sides[math]\theta = 45°[/math]
Pre Calc Question: sintheta+costheta=square root of 2?
I'm going to say x = theta to make it a little easier to write. We have: sinx + cosx = sqrt(2) First, square both sides, remembering to FOIL the left side: sin^2x + 2sinxcosx + cos^2x = 2 Re-arrange the left: sin^2x + cos^2x + 2sincosx = 2 The Pythagorean Identity tells us sin^2x + cos^2x = 1, and so we can substitute: 1 + 2sinxcosx = 2 Subtract 1 from each side: 2sinxcosx = 1 Divide by 2: sinxcosx = 1/2 Now, use the product-to-sum formula to convert the left side to: 1/2 [sin(x + x) + sin(x - x)] = 1/2 Multiply both sides by 2, and group the x terms: sin(2x) + sin(0) = 1 We know sin(0) = 0, and so: sin(2x) = 1 Take the arcsin of each side: 2x = arcsin(1), which is: 2x = pi/2, 5pi/2, etc Divide by 2 to get the solutions: x, or theta = pi/4, 5pi/4 on the domain [0, 2pi).