MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Basic Trigonometric Identities

Prove by vect...

Question

Prove by vector method: $sin (α - β) = sin α \cdot cos β - cos α \cdot sin β$ .

Open in App

Solution

$sin (α - β) = sin α cos β - cos α sin β$ .
Let $^i$ is a vector along $X$ -axis $^j$ is a unit vector along $Y$ -axis and
$∠ X O P = α$
$∠ X O Q = β$
$∴ ∠ Q O P = (α - β)$
$\to O M = x_{1}^i$ $\to M P = y_{1}^j$
In $Δ$ OPM $\to O P = \to O M + \to M P$
$\to O P = x_{1}^i + y_{1}^j$
$cos α = \frac{O M}{O P} = \frac{x_{1}}{1}$
$sin α = \frac{M P}{O P} = \frac{y_{1}}{1}$
$\to O P =^i cos α +^j sin α$
Similarly $\to O Q =^i cos β +^j sin β$
$\to O Q \times \to O P = (^i cos β +^j sin β) \times (^i cos α +^j sin α)$
$\to O Q \times \to O P = sin α cos β (^i \times^j) + cos α sin β (^j \times^i)$
$\to O Q \times \to O P = (sin α cos β - cos α sin β)^k$ .......(i)
$\to O Q \times \to O P = | O Q | | O P | sin (α - β)^k$
$= ∣ ∣ ∣ ∣ \begin{matrix} i & j & k cos β & sin β & o cos α & sin α & 0 \end{matrix} ∣ ∣ ∣ ∣$
$\to O Q \times \to O P = sin (α - β)^k$ .......(ii)
$∴$ By equation (i) and (ii)
$sin (α - β) = sin α cos β - cos α sin β$ .

Suggest Corrections

thumbs-up

0

Similar questions

Q. Prove by vector method that

sin (α + β) = sin α cos β + cos α sin β

Q. Using vector, prove that

sin (α - β) = sin α cos β - cos α sin β

Q. Prove by using vectors that

sin (α + β) = sin α cos β + cos α sin β

Q. Prove that: If

0 < α, β, γ < \frac{π}{2}

sin α + sin β + sin γ > sin (α + β + γ)

\frac{sin (α + β)}{sin (α - β)} = \frac{a + b}{a - b}

then prove that

a tan β = b tan α

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Pythagorean Identities

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Basic Trigonometric Identities

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon