If cos - - -1 and cos - - -12- where - - - - then the number of ordered pairs - - satisfying both the equations is

Cos (α β) =1 ⇒α β=2nπ, n∈ℤ Since nbsp;α, β∈ ( π, π), we have α β =0 ⇒α =β Now, nbsp;cos (α + β)=12 nbsp; ⇒cos 2 α =12 Clearly, y=cos 2x =12 has four ...

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

Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Common Angles

If cos α -β =...

Question

If $cos (α - β) = 1$ and $cos (α + β) = \frac{1}{2},$ where $α, β \in (- π, π),$ then the number of ordered pairs $(α, β)$ satisfying both the equations is

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $4$
$cos (α - β) = 1$
$\Rightarrow α - β = 2 n π, n \in Z$
Since $α, β \in (- π, π),$ we have
$α - β = 0$
$\Rightarrow α = β$

Now, $cos (α + β) = \frac{1}{2}$
$\Rightarrow cos 2 α = \frac{1}{2}$

Clearly, $y = cos 2 x = \frac{1}{2}$ has four solutions in $(- π, π)$

Suggest Corrections

Similar questions

c o s (α - β) = 1

and

c o s (α + β) = \frac{1}{e}

, where

α, β \in [- π, π]

. Number of pairs of

α, β

which satisfy both the equation is

Q. The number of ordered pairs

(α, β)

, where

α, β

\in

(- π, π)

satisfying

cos (α - β) = 1

and

cos (α + β) = \frac{1}{e}

Q. If

cos (α - β) = 1

and

cos (α + β) = \frac{1}{e}

, where

α, β \in [- π, π]

. Then number of ordered pairs of

(α, β)

c o s (α - β) = 1

and

c o s (α + β) = \frac{1}{e}

, where

α, β \in [- π, π]

. Number of pairs of

α, β

which satisfy both the equation is

Q. If

cos (α - β) = 1

and

cos (α + β) = \frac{1}{e}

, where

- π < α, β < π

, then total number of ordered pair of

(α, β)

Join BYJU'S Learning Program

If cos(α−β)=1 and cos(α+β)=12, where α,β∈(−π,π), then the number of ordered pairs (α,β) satisfying both the equations is

The correct option is D 4cos(α−β)=1 ⇒α−β=2nπ,n∈Z Since α,β∈(−π,π), we have α−β=0 ⇒α=β Now, cos(α+β)=12 ⇒cos2α=12 Clearly, y=cos2x=12 has four solutions in (−π,π)

If $cos (α - β) = 1$ and $cos (α + β) = \frac{1}{2},$ where $α, β \in (- π, π),$ then the number of ordered pairs $(α, β)$ satisfying both the equations is