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The expression (\cos^2(90^\circ - \theta)) can be simplified using the co-function identity, which states that (\cos(90^\circ - \theta) = \sin(\theta)). Therefore, (\cos^2(90^\circ - \theta) = \sin^2(\theta)). This means that (\cos^2(90^\circ - \theta)) is equal to the square of the sine of (\theta).
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Sin squared theta plus cos squared theta barabar Kya Hota Hai?
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How do you solve cos theta subtract cos squared theta divide 1 plus cos theta?
The question contains an expression but not an equation. An expression cannot be solved.
How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
How do you solve 4 cosine squared theta equals 1?
4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = ±1/2 Now cos(theta) = 1/2 => theta = 60 + 360k or theta = 300 + 360k while Now cos(theta) = -1/2 => theta = 120 + 360k or theta = 240 + 360k where k is an integer.
What is the identity for tan theta?
The identity for tan(theta) is sin(theta)/cos(theta).
What is cos theta times cos theta?
Cos theta squared
What is sec squared theta times cos squared theta minus tan squared theta?
Tan^2
What is cos 90 minus theta?
cosine (90- theta) = sine (theta)
What does negative sine squared plus cosine squared equal?
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
Sin squared theta plus cos squared theta barabar Kya Hota Hai?
1
How do you solve theta if cos squared theta equals 1 and 0 is less than or equal to theta which is less than 2pi?
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
How do you solve cos theta subtract cos squared theta divide 1 plus cos theta?
The question contains an expression but not an equation. An expression cannot be solved.
How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Which basic trigonometric identity is actually a statement of the pythagorean theorem?
COS squared Theta + SIN squared Theta = 1; where Theta is the angles measurement in degrees.
What is cos theta minus cos theta times sin squared theta?
cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)
How would you solve and show work for cos2 theta if cos squared theta equals 1 and theta is in the 4th quadrant?
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
How do you integrate cos squared theta times sine theta?
To integrate ( \cos^2 \theta \sin \theta ), you can use a substitution method. Let ( u = \cos \theta ), then ( du = -\sin \theta , d\theta ). The integral becomes ( -\int u^2 , du ), which evaluates to ( -\frac{u^3}{3} + C ). Substituting back, the final result is ( -\frac{\cos^3 \theta}{3} + C ).
