Question

If sinβ is the geometric mean between sinα, cosα then cos2βisequal to

Answer 1

first sin is sin beta then sin alpha,cos alpha,cos 2beta

Answer 2

OK.\[\sin^2 \beta = \sin\alpha \cdot \cos\alpha\]And we know that,\[\cos(2\beta) = 1 - 2\sin^2(\beta)\]

Answer 3

\[if \sin \beta,is the geo mean btw \sin \alpha,\cos \alpha,then \cos 2\]

Answer 4

\[\beta 2 \iota [ 3 x 4 ] \]

Answer 5

@Rahul .cant able to follow

Answer 6

Do you understand my solution?

Answer 7

thats what i have studided

Answer 8

sin beta=sin alpha cos alpha

Answer 9

yes

Answer 10

ParthKohli you are wrong in one thing

Answer 11

If \(b\) is the geometric mean between \(a\) and \(c\), it means that \(b^2 = ac\).

By the above property, it can be said that \(\sin^2 \beta = \sin\alpha \cdot \cos \alpha\).

We need to determine the value of \(\cos(2\beta)\) which is equal to \(1- 2 \sin^2 \beta \). Plug in the value of \(\sin^2 \beta.\)

Answer 12

\[1-2\sin \alpha \cos\]

Answer 13

That's right. You can simplify it further.

Answer 14

1-2 sin alpha cos alpha

Answer 15

Parth Kohli you know what

Answer 16

1-sin 2 alpha

Answer 17

That's right!\[\therefore \cos(2\beta) = 1 - \sin(2\alpha)\]

Answer 18

ParthKohli more like Virat Kohli the cricketer

Answer 19

lol

Answer 20

common ler become fan of each other

Answer 21

then

Answer 22

alpha can be divided

Answer 23

how

Answer 24

@ParthKohli

Answer 25

if you get the divisor then because the dividend is alpha

Answer 26

how

Answer 27

you have to get the divisor by its formula

Answer 28

there must be a formula

Answer 29

\[If \sin \beta is the geometric mean \between \sin \alpha,\cos \alpha then \cos 2\beta is equal \to\]

Answer 30

I've already given you the answer...

Answer 31

@ParthKohli the problem didnt get finished

Answer 32

see question 29

Answer 33

oh u live in india !

Answer 34

seeing the question

Answer 35

hmm..lol

Answer 36

question taken from online exam.i was searching through net

Answer 37

continuing parth's wrok :
\[\large \begin{array}\\ 1-\sin 2\alpha &= \sin^2\alpha + \cos^2\alpha - 2\sin\alpha \cos\alpha \\&= (\sin \alpha - \cos \alpha)^2 \end{array} \]

Answer 38

\[\large \begin{array}\\ &= \color{red}{2}(\sin \alpha \color{red}{\cos(\pi/4) }- \cos \alpha\color{red}{\sin(\pi/4)})^2 \end{array} \]

Answer 39

\[\large \begin{array}\\ &= \color{red}{2}(\sin (\alpha-\pi/4) )^2\end{array} \]

\[\large \begin{array}\\ &= \color{red}{2}\sin^2 (\pi/4-\alpha )\end{array} \]