Trigonometry+ 10: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 10


Problem A


Prove that ∠BAC = ½∠AOC.
Using this, derive the trigonometric identity for any acute angle α:
sin(α)·tan(½α) = 1−cos(α)


Problem B

Solve the equation
arcsin(x) = arccos(x)

Solution B
Let's recall the domain, co-domain and draw a graph of both functions.
Domain for both arcsin(x) and arccos(x) is x∈[−1,1].
Co-domain for arcsin(x) is x∈[−π/2,π/2].
Co-domain for arccos(x) is x∈[0,π].
Below are graphs of functions arcsin(x) (red) and arccos(x) (blue). This graph will be used in the next problem as well, that's why we marked points x=1/4 and x=11/16.

An obvious guess of the solution to the above equation, where both graphs intersect, would be x=√2/2 since, as we all know,
sin(π/4)=cos(π/4)=√2/2
and, therefore,
arcsin(√2/2)=π/4 and
arccos(√2/2)=π/4.
An easy proof to this is as follows.
Take a sine function of both parts of an original equation
sin(arcsin(x)) = sin(arccos(x))
x = sin(arccos(x))
The value of x must be positive, as we see from the graph above.
Therefore,
x = √1−cos²(arccos(x))
x = √1−x²
x² = 1 − x²
x² = 1/2
(retaining only positive solution)
x = √2/2


Problem C

Calculate
3·arcsin(1/4) + arccos(11/16)

Solution C

Let's set for brevity
α = arcsin(1/4)
⇒ sin(α) = 1/4
β = arccos(11/16)
⇒ cos(β) = 11/16

Considering the positive values of arcsin()'s and arccos()'s arguments, the angles α and β they represent must be in the first quarter between 0 and π/2.

From sin²(α)+cos²(α)=1 and
sin(α)=1/4 follows:
cos(α)=√15/4

Since cosβ=11/16, then
sin(β)=3√15/16

Also, it looks like we will need formulas for a sine and a cosine of a triple angle.
Let's derive them.
sin(3x) = sin(x+2x) =
= sin(x)·cos(2x) +
+ cos(x)·sin(2x) =
= sin(x)·[1−2sin²(x)] +
+ cos(x)·2sin(x)·cos(x) =
=sin(x) − 2sin³(x) +
+ 2sin(x)·[1−sin²(x)] =
= 3sin(x) − 4sin³(x)

Using these formulas, we can evaluate a sine of the sum of two angles in this problem.
sin[3·arcsin(1/4) +
+ arccos(11/16)] =
= sin(3α + β) =
= sin(3α)·cos(β)+cos(3α)·sin(β)

Evaluating the components of the above expression separately:

sin(3α) = 3sin(α) − 4sin³(α) =
= 3/4 − 4/64 = 11/16

cos(β) = 11/16

cos(3α) = √1−121/256 =
= 3√15/16

sin(β) = 3√15/16

Combining all these values, we obtain
sin(3α + β) =
= 121/256 + 135/256 = 1

Since a sine of an angle, that is a sum of two acute angles, is equal to 1, the angle itself is equal to π/2.