Question

Use a double angle formula the expression
a)sinxcosx
b)cos^2(x)- 1/2
c) 10sin^2(x)-5
d) (sinx-cosx)(sinx-cosx)

Answer 1

*rewrite the expression

Answer 2

a) use sin2x = 2sinxcosx

Answer 3

b) use cos2x = 2cos^2x - 1

Answer 4

recall that -> sin(2x)=2sin(x)cos(x)

so what you'd do is, make the right-hand-side look like sin(x)cos(x)
by simplifying

so what do you think you could use to make \(\bf {\color{blue}{ sin(2x)}}=2sin(x)cos(x)\implies {\color{blue}{ \square ?}}=sin(x)cos(x)\quad ?\)

Answer 5

2

Answer 6

times 2? over 2? under 2? + 2?

Answer 7

times 2...maybe?

Answer 8

let's see that \(\bf {\color{blue}{ sin(2x)}}=2sin(x)cos(x)\implies 2{\color{blue}{ sin(2x)}}=2\cdot 2sin(x)cos(x)
\\ \quad \\
\implies 2{\color{blue}{ sin(2x)}}=4sin(x)cos(x)\)

nope, the left-side doesn't look like sin(x)cos(x)

Answer 9

right-hand-side rather

Answer 10

so times 1?

Answer 11

\(\bf {\color{blue}{ sin(2x)}}=2sin(x)cos(x)\implies 1{\color{blue}{ sin(2x)}}=1\cdot 2sin(x)cos(x)
\\ \quad \\
\implies 2{\color{blue}{ sin(2x)}}=2sin(x)cos(x)\) no dice on that either, the right-side doesn't look like sin(x)cos(x) yet

Answer 12

okay, I give up

Answer 13

woops, darn typos... anyhow \(\bf {\color{blue}{ sin(2x)}}=2sin(x)cos(x)\implies 1{\color{blue}{ sin(2x)}}=1\cdot 2sin(x)cos(x)
\\ \quad \\
\implies {\color{blue}{ sin(2x)}}=2sin(x)cos(x)\)
still though, no dice on that one

Answer 14

lemme rewrite it a bit differently

Answer 15

actually... lemme show it instead...
let use DIVIDE by 2 both sides

\(\bf {\color{blue}{ sin(2x)}}=2sin(x)cos(x)\implies \cfrac{{\color{blue}{ sin(2x)}}}{2}=\cfrac{\cancel{2}sin(x)cos(x)}{\cancel{2}}\) what do you think?

Answer 16

i'm following

Answer 17

well, what would the right-side give you?

Answer 18

sin(x)cos(x) right?

Answer 19

\(\bf {\color{blue}{ sin(2x)}}=2sin(x)cos(x)\implies \cfrac{{\color{blue}{ sin(2x)}}}{2}=\cfrac{\cancel{2}sin(x)cos(x)}{\cancel{2}}\\ \quad \\
\implies \cfrac{{\color{blue}{ sin(2x)}}}{2}=sin(x)cos(x)\)

Answer 20

okay so what about b?

Answer 21

the b) one, as cwrw238 already pointed out \(\bf {\color{blue}{ cos(2x)}}=2cos^2(x)-1\implies {\color{blue}{ \square ?}}=cos^2(x)-\frac{1}{2}\)

so you'd simplify till the right-hand-side looks like that :)

try also dividing both sides by 2, that is \(\bf \cfrac{{\color{blue}{ cos(2x)}}}{2}=\cfrac{2cos^2(x)-1}{2}\)

Answer 22

You're a life-saver thank you

Answer 23

wait, what about d?