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If there is a plus in between, that would be equal to 1, as a result of the Pythagorean Theorem. Otherwise, you can convert this into other forms with some of the trigonometric identities for multiplication, but you won't really get it into a simpler form.
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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.
Sin squared theta plus cos squared theta barabar Kya Hota Hai?
1
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
If a cos theta plus b sin theta equals 8 and a sin theta - b cos theta equals 5 show that a squared plus b squared equals 89?
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
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).
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)
What is sec squared theta times cos squared theta minus tan squared theta?
Tan^2
