SOLUTION: sec theda - cos theda = sin thedacos theda
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Question 358056: sec theda - cos theda = sin thedacos theda
Found 3 solutions by Fombitz, stanbon, nyc_function:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Minor correction: It's theta not theda.
What's your question though?
.
.
.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
sec theda - cos theda = sin thedacos theda
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sec - cos = sin*cos
---
(1/cos) - cos = sin*cos
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(1-cos^2)/cos = sin*cos
----
sin^2/cos = sin*cos
---
sin^2 = sin(cos^2)
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The original equation is not an identity.
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There is no value of theda that will make
the original equation true.
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Cheers,
Stan H.
Answer by nyc_function(2741) (Show Source): You can put this solution on YOUR website!
The word is theta not theda.
You did not say whay exactly needs to be done here but my guess is proving trig identities.
When proving trig identities, convert everything to sine and cosine.
secθ - cosθ = sinθcosθ
We know that secθ = 1/cosθ.
So, replace sine theta with 1/cos theta.
1/cosθ - cosθ = sinθcosθ
On the left side, we simply apply the rules for subtraction of fractions.
The left side becomes (1 - cos^2θ)/cosθ.
The equation now looks like this:
(1 - cos^2θ)/cosθ = sinθcosθ
Can you take it from here?
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