Unizor – Creative Mind through Art of Mathematics: Trigonometry+ 10: UNIZOR.COM – Math+ & Problems

Notes to a video lecture on http://www.unizor.com Trigonometry+ 10Problem AProve 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 equationarcsin(x) = arccos(x)Solution BLet’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)=√2/2and, therefore,arcsin(√2/2)=π/4 andarccos(√2/2)=π/4.An easy proof to this is as follows.Take a sine function of both parts of an original equationsin(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−x²x² = 1 − x²x² = 1/2(retaining only positive solution)x = √2/2Problem CCalculate3·arcsin(1/4) + arccos(11/16)Solution CLet’s set for brevityα = arcsin(1/4)⇒ sin(α) = 1/4β = arccos(11/16)⇒ cos(β) = 11/16Considering 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 andsin(α)=1/4 follows:cos(α)=√15/4Since cosβ=11/16, thensin(β)=3√15/16Also, 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 = 11/16cos(β) = 11/16cos(3α) = √1−121/256 == 3√15/16sin(β) = 3√15/16Combining all these values, we obtainsin(3α + β) == 121/256 + 135/256 = 1Since 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.