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 !
Answer:
49
Step-by-step explanation:
- Try finding pattern of last two digits of 7 to the power of some numbers.
7^1=07
7^2=49
7^3=343
7^4= 2301
7^5= 16807
- so , the pattern is, 07,49,43,01
- so, if we start writing all numbers from 1 to 2018 , four number in each line, 2018 will fall in the second column.
- so it will have 49 as the last 2 digits [by seeing this pattern : 07,49,43,01 ]
Its A. x=4
procedure: 3*equation1 - 2*equation2 ⇒ x=4
The answer to the question
Answer:
B i think
Step-by-step explanation:
