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What else can I help you with?
How do you simplify tan theta cos theta?
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
What is tan squared theta minus sec squared theta simplified?
tan θ = sin θ / cos θ sec θ = 1 / cos θ sin ² θ + cos² θ = 1 → sin² θ - 1 = - cos² θ → tan² θ - sec² θ = (sin θ / cos θ)² - (1 / cos θ)² = sin² θ / cos² θ - 1 / cos² θ = (sin² θ - 1) / cos² θ = - cos² θ / cos² θ = -1
What is sin theta cos theta?
It's 1/2 of sin(2 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
Is sin squared theta and sin theta squared the same?
Your question is insufficiently precise, but I'll try to answer anyway. "Sine squared theta" usually means "the value of the sine of theta, quantity squared". "Sine theta squared" usually means "the value of the sine of the quantity theta*theta". The two are not at all the same.
If cos and theta 0.65 what is the value of sin and theta?
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
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?
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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 cos theta times cos theta?
Cos theta squared
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.
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 simplify tan theta cos theta?
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
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 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 ).
De Morgan's law in complex number?
(Sin theta + cos theta)^n= sin n theta + cos n theta
What is the identity for tan theta?
The identity for tan(theta) is sin(theta)/cos(theta).
