Show that ( 2 + 3 ) sin θ + 2 cos θ lies between − ( 2 + 5 ) and ( 2 + 5 ) .

We know that a sin β + b cos β ∈ - a 2 + b 2 , a 2 + b 2 and the range of sin x & cos x is - 1 , 1 Let, A = ( 2 + 3 ) sin θ + 2 cos θ ⇒ − ( 2 + 3 ) 2 + 2 2 ≤ A ≤ ( 2 + 3 ) 2 + 2 2 ⇒ − 11 + 4 3 ≤ A ≤ 11 + 4 3 Here, 11 + 4 3 ≈ 17 . 928 ⇒ 11 + 4 3 ≈ 4 . 328 ⇒ 11 + 4 3 ≈ 2 + 5 So, the range of ( 2 + 3 ) sin θ + 2 cos θ is approximately lies between - ( 2 + 5 ) and ( 2 + 5 ) .

Show that ( 2 + 3 ) sin θ + 2 cos θ lies between − ( 2 + 5 ) and ( 2 + 5 ) .

We know that a sin β + b cos β ∈ - a 2 + b 2 , a 2 + b 2 and the range of sin x & cos x is - 1 , 1 Let, A = ( 2 + 3 ) sin θ + 2 cos θ ⇒ − ( 2 + 3 ) 2 + 2 2 ≤ A ≤ ( 2 + 3 ) 2 + 2 2 ⇒ − 11 + 4 3 ≤ A ≤ 11 + 4 3 Here, 11 + 4 3 ≈ 17 . 928 ⇒ 11 + 4 3 ≈ 4 . 328 ⇒ 11 + 4 3 ≈ 2 + 5 So, the range of ( 2 + 3 ) sin θ + 2 cos θ is approximately lies between - ( 2 + 5 ) and ( 2 + 5 ) .

Show that 5 cos θ + 3 cos ( θ + π 3 ) + 3 lies between - 4 and 10 .

Let A = 5 cos θ + 3 cos ( θ + π 3 ) + 3 A = 5 cos θ + 3 ( cos θ 2 − 3 sin θ 2 ) + 3 A = 13 2 cos θ − 3 3 2 sin θ + 3 . . . . . . . 1 We know − A 2 + B 2 ≤ A cos θ − B sin θ ≤ A 2 + B 2 ⇒ - ( 13 2 ) 2 + ( 3 3 2 ) 2 ≤ 13 2 cos θ − 3 3 2 sin θ ≤ ( 13 2 ) 2 + ( 3 3 2 ) 2 ⇒ − 196 4 ≤ 13 2 cos θ - 3 3 2 sin θ ≤ 196 4 ⇒ − 7 ≤ 13 2 cos θ − 3 3 2 sin θ ≤ 7 Putting in 1 , we get − 7 + 3 ≤ A ≤ 7 + 3 − 4 ≤ A ≤ 10 .

Show that the minimum value of sin 8 x + cos 8 x is 1 8 .

Let A = sin 8 x + cos 8 x Here, sin 8 x ≥ 0 and cos 8 x ≥ 0 For two non-negative numbers sin 8 x & cos 8 x , applying A . M . ≥ G . M . ⇒ sin 8 x + cos 8 x 2 ≥ sin 8 x cos 8 x ⇒ A ≥ 2 sin 4 x . cos 4 x ⇒ A ≥ 2 ( sin x cos x ) 4 ⇒ A ≥ 2 sin 2 x 2 4 ⇒ A ≥ 2 16 sin 4 2 x ⇒ A ≥ 1 8 , as sin 4 2 x ∈ 0 , 1 Hence, sin 8 x + cos 8 x ≥ 1 8

Show that sin 2 n x + cos 2 n x ≤ 1 .

Consider the identity, sin 2 x + cos 2 x = 1 . . . . i Since, − 1 ≤ sin x ≤ 1 ⇒ 0 ≤ sin 2 x ≤ 1 ⇒ sin 2 x ≥ sin 4 x ≥ sin 6 x ≥ sin 8 x . .. ≥ sin 2 n x . . . . i i As fraction multiplied by a fraction will always be less than both the fractions and equality will hold in above case when sin x = 0 , 1 , - 1 Also, − 1 ≤ cos x ≤ 1 ⇒ 0 ≤ cos 2 x ≤ 1 ⇒ cos 2 x ≥ cos 4 x ≥ cos 6 x ≥ cos 8 x . .. ≥ cos 2 n x . . . . i i i As fraction multiplied by a fraction will always be less than both the fractions and equality will hold in above case when cos x = 0 , 1 , - 1 From i , i i & i i i , sin 2 x + cos 2 x ≥ sin 2 n x + cos 2 n x ⇒ sin 2 n x + cos 2 n x ≤ 1 Hence, Proved.

If 0 < α < π 2 then show that tan α + cot α > sin α + cos α .

For the given interval, 0 < α < π 2 , both tan α and cot α are positive. For two positive numbers, A . M . ≥ G . M . ⇒ tan α + cot α 2 ≥ tan α . cot α ⇒ tan α + cot α 2 ≥ sin α cos α . cos α sin α ⇒ tan α + cot α ≥ 2 . . . i Consider, sin α + cos α = 2 1 2 sin α + 1 2 cos α ⇒ sin α + cos α = 2 cos π 4 sin α + sin π 4 cos α ⇒ sin α + cos α = 2 sin ( π 4 + α ) . . . . i i Also, - 1 ≤ sin π 4 + α ≤ 1 . . . . i i i From i i , - 2 ≤ sin α + cos α ≤ 2 , using i i i From i & i i i , tan α + cot α ≥ sin α + cos α Hence, Proved.

If α > 0 , β > 0 , γ > 0 and α + β + γ = π 2 . Show that tan 2 α + tan 2 β + tan 2 γ ≥ 1 .

Given, α + β + γ = π 2 ⇒ α + β = π 2 − γ . . . . i ⇒ tan ( α + β ) = tan π 2 - γ ⇒ tan ( α + β ) = cot γ ⇒ tan α + tan β 1 − tan α · tan β = 1 tan γ ⇒ tan α tan γ + tan β tan γ + tan α tan β = 1 . . . . i i Since, α , β , γ > 0 & α + β + γ = π 2 ⇒ α , β , γ ∈ 0 , π 2 So, applying A . M . ≥ G . M . on two positive quantities tan 2 α & tan 2 β tan 2 α + tan 2 β 2 ≥ tan 2 α . tan 2 β ⇒ tan 2 α + tan 2 β ≥ 2 tan α tan β . . . . i i i Similarly below inequalities can also be considered, tan 2 β + tan 2 γ ≥ 2 tan β tan γ . . . . i v tan 2 γ + tan 2 α ≥ 2 tan γ tan α . . . . v Now, i i i + i v + v gives, tan 2 α + tan 2 β + tan 2 γ ≥ tan α tan β + tan β tan γ + tan γ tan α ⇒ tan 2 α + tan 2 β + tan 2 γ ≥ 1 , using i i , tan α tan γ + tan β tan γ + tan α tan β = 1 . Hence, Proved.

Prove that 0 ≤ cos 2 α + cos 2 ( α + β ) − 2 cos α cos β cos ( α + β ) ≤ 1 .

Consider, cos 2 α + cos 2 ( α + β ) − 2 cos α cos β cos ( α + β ) = cos 2 α + cos 2 ( α + β ) − [ cos ( α + β ) + cos ( α − β ) ] cos ( α + β ) = cos 2 α − cos ( α − β ) cos ( α + β ) = cos 2 α − ( cos 2 α − sin 2 β ) = sin 2 β Since, sin β ∈ - 1 , 1 ⇒ sin 2 β ∈ 0 , 1 So, cos 2 α + cos 2 ( α + β ) − 2 cos α cos β cos ( α + β ) = sin 2 β ∈ 0 , 1 Hence, Proved.