Let's work on the left side first. And remember that
the<u> tangent</u> is the same as <u>sin/cos</u>.
sin(a) cos(a) tan(a)
Substitute for the tangent:
[ sin(a) cos(a) ] [ sin(a)/cos(a) ]
Cancel the cos(a) from the top and bottom, and you're left with
[ sin(a) ] . . . . . [ sin(a) ] which is [ <u>sin²(a)</u> ] That's the <u>left side</u>.
Now, work on the right side:
[ 1 - cos(a) ] [ 1 + cos(a) ]
Multiply that all out, using FOIL:
[ 1 + cos(a) - cos(a) - cos²(a) ]
= [ <u>1 - cos²(a)</u> ] That's the <u>right side</u>.
Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.
So, on the <u>right side</u>, you could write [ <u>sin²(a)</u> ] .
Now look back about 9 lines, and compare that to the result we got for the <u>left side</u> .
They look quite similar. In fact, they're identical. And so the identity is proven.
Whew !
This would be 4y + 1ytytyty I believe considering they’re all bunched together in the last bit.
Answer:
ok
Step-by-step explanation:
Answer:
(14-9i)+(2+3i) = 16-6i
Step-by-step explanation:
We need to find the equivalent of (14-9i)+(2+3i).
We need to add the two expressions.
(14-9i)+(2+3i) = 14+2 -9i+3i
= 16-6i
Here, the real part is 16
The imaginary part is -6
Hence, the required answer is equal to 16-6i.
