sin(<em>θ</em>) + cos(<em>θ</em>) = 1
Divide both sides by √2:
1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = 1/√2
We do this because sin(<em>x</em>) = cos(<em>x</em>) = 1/√2 for <em>x</em> = <em>π</em>/4, and this lets us condense the left side using either of the following angle sum identities:
sin(<em>x</em> + <em>y</em>) = sin(<em>x</em>) cos(<em>y</em>) + cos(<em>x</em>) sin(<em>y</em>)
cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)
Depending on which identity you choose, we get either
1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = sin(<em>θ</em> + <em>π</em>/4)
or
1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = cos(<em>θ</em> - <em>π</em>/4)
Let's stick with the first equation, so that
sin(<em>θ</em> + <em>π</em>/4) = 1/√2
<em>θ</em> + <em>π</em>/4 = <em>π</em>/4 + 2<em>nπ</em> <u>or</u> <em>θ</em> + <em>π</em>/4 = 3<em>π</em>/4 + 2<em>nπ</em>
(where <em>n</em> is any integer)
<em>θ</em> = 2<em>nπ</em> <u>or</u> <em>θ</em> = <em>π</em>/2 + 2<em>nπ</em>
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We get only one solution from the second solution set in the interval 0 < <em>θ</em> < 2<em>π</em> when <em>n</em> = 0, which gives <em>θ</em> = <em>π</em>/2.
Answer:
(x-5)(x+1)
Step-by-step explanation:
Since the highest power of x is 2, u can assume the equation to be (x +/- A) (x +/- B), whereby A and B are unknowns.
Since the coefficient of X^2 is 1, and the number in the equation is - 5, the only possible values for A and B to get - 5 is either - 5 and 1 or 1 and - 5.
Since coefficient of x is - 4 and number in equation is - 5, imagine inserting those numbers in the equation and cross multiplying back. You would realise that the answer should be as listed above.
Answer:
5,280 times
hope this helps
have a good day :)
Step-by-step explanation:
I hated doing that stuff, it was a pain for me
I can’t see the image can you post it please or just tell me what the problem is