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cos(A + B) = cosAcosB - sinAsinB, with A=B, cos(2A) = cos2A - sin2A, then you can use cos2A + sin2A = 1, to produce more, like: [2cos2A - 1] or [1 - 2sin2A], and others.

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What does negative sine squared plus cosine squared equal?

To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)


If cos and theta 0.65 what is the value of sin and theta?

Remember use the Pythagorean Trig/ Identity. Sin^(2)(Theta) + Cos^(2)(Theta) = 1 Algebraically rearrange Sin^(2)(Theta) = 1 - Cos^(2)(Theta) Substitute Sin^(2)(Theta) = 1 - 0.65^(2) Factor Sin^(2)(Theta) = ( 1- 0.65 )( 1 + 0.65) Sin^(2)(Theta) = (0.35)(1.65) Sin^(2)(Theta) = 0.5775 Sin(Theta) = sqrt(0.5775) Sin(Theta) = 0.759934207.... Theta = Sun^(-1)(0.759934207...) Theta = 49.45839813 degrees.


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 ).


How do you finish the equation sin 2 theta?

(/) = theta sin 2(/) = 2sin(/)cos(/)


What is sec squared theta times cos squared theta minus tan squared theta?

Sec^(2) = 1 / Cos^(2) Hence Sec^(2) X Cos^(2) = 1/Cos^(2) X Cos^(2) Cancel down by ' Cos^(2) '. 1/1 X 1 = 1 1 - Tan^(2) (The ANSWER) !!!!! Factor ( 1 - Tan)(1 + Tan)

Related Questions

What does negative sine squared plus cosine squared equal?

To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)


What is cos squared 90 - theta?

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).


What is sec theta - 1 over sec theta?

Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta


What is sin theta cos theta?

It's 1/2 of sin(2 theta) .


What is cos theta times cos theta?

Cos(Theta) X Cos(Theta) = Cos^(2)[Theta].


If cos and theta 0.65 what is the value of sin and theta?

Remember use the Pythagorean Trig/ Identity. Sin^(2)(Theta) + Cos^(2)(Theta) = 1 Algebraically rearrange Sin^(2)(Theta) = 1 - Cos^(2)(Theta) Substitute Sin^(2)(Theta) = 1 - 0.65^(2) Factor Sin^(2)(Theta) = ( 1- 0.65 )( 1 + 0.65) Sin^(2)(Theta) = (0.35)(1.65) Sin^(2)(Theta) = 0.5775 Sin(Theta) = sqrt(0.5775) Sin(Theta) = 0.759934207.... Theta = Sun^(-1)(0.759934207...) Theta = 49.45839813 degrees.


What divided by cosine squared theta equals one?

The equation that satisfies the condition "what divided by cosine squared theta equals one" is simply the expression itself. If we let ( x ) be the quantity, then the equation can be expressed as ( \frac{x}{\cos^2 \theta} = 1 ). Solving for ( x ) gives ( x = \cos^2 \theta ). Thus, ( \cos^2 \theta ) divided by ( \cos^2 \theta ) equals one.


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 ).


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.


How do you finish the equation sin 2 theta?

(/) = theta sin 2(/) = 2sin(/)cos(/)


How do you simplify cos theta times csc theta divided by tan theta?

'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2


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