Question

evaluate Cos48°-Sin42°​

Answer 1

To find Cos 48o – Sin 42o

We know that Cos (90 – θ) = Sin θ……..(1)

So we can write Cos 48o as

Cos 48o = Cos (90 – 42)

Cos 48o = Cos (90 – 42) = Sin 42o………(2)

Cos 48o – Sin 42o

From equation (2) we can write the above equation as

Sin 42o – Sin 42o

= 0

Answer 2

Topic:-

Trigonometry

Question:-

Cos48° - Sin42° = ?

Solution:-

Cos48° - Sin42°

We know that Cos 48°= Cos(90-42)

Cos(90-42) = Sin42°

Now we have to substitute values

Sin42°- Sin42°

=0

Answer:-

Cos48° - Sin42° = 0

More information :-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj