Trigonometric functions (Mathematics II)

To convert degrees to radians, multiply by π/180: 60×π/180 = π/3. The value π/6 is 30°, π/4 is 45° and 2π/3 is 120°. Taking 180°=π as the reference, it is quicker to note that 60° is one third of 180°, hence π/3.

To convert radians to degrees, multiply by 180/π: (5π/6)×(180/π) = 5×30 = 150°. You can also see it as five times π/6=30°. The value 210° confuses this with 7π/6, and 30° is just π/6.

In radian measure the arc length is l = rθ = 6×π/3 = 2π. The option 6π computes the area (1/2)r²θ instead, and 12π is r²θ without the factor one half. Do not mix up the formulas for arc length and area.

S = (1/2)r²θ = (1/2)×36×(π/3) = 6π. The option 12π forgets the factor one half, and 2π confuses this with the arc length rθ. The form S=(1/2)rl using the arc length also gives (1/2)×6×2π=6π.

Since 7π/6 = π+π/6, the terminal ray lies in the third quadrant, where sin, the y-coordinate, is negative. Its size is sin(π/6)=1/2, so the value is −1/2. The option −√3/2 confuses this with the cosine side, and 1/2 omits the sign.

Since 4π/3 = π+π/3, this is in the third quadrant, where tan is positive because sin and cos are both negative and their ratio is positive. So the value is tan(π/3) = √3. The option −√3 gets the sign wrong, and 1/√3 confuses this with tan(π/6).

The period of y=sin kθ is 2π/k. With k=2 it is 2π/2 = π. The rotation is twice as fast, so the wave is compressed to half its length. The value 2π is the period of sinθ itself, and 4π comes from multiplying 2π by k.

The period is 2π/k with k=1/2, so it is 2π÷(1/2) = 4π. The rotation is half as fast, so the wave is stretched to twice its length. The value π comes from multiplying 2π by 1/2 instead.

From sin²θ+cos²θ=1 we get cos²θ = 1−9/25 = 16/25, so cosθ = ±4/5. In the second quadrant cos is negative, so cosθ = −4/5. The option 4/5 omits the quadrant sign, and −3/4 confuses this with tan.

sinθcosθ = sinθcosθ/(sin²θ+cos²θ) = tanθ/(1+tan²θ) = 2/(1+4) = 2/5. In the third quadrant sin and cos are both negative, so their product is positive, which matches the sign. The option −2/5 mechanically makes it negative because of the quadrant.

sin75° = sin(45°+30°) = sin45°cos30°+cos45°sin30° = (√2/2)(√3/2)+(√2/2)(1/2) = (√6+√2)/4. The option (√6−√2)/4 is sin15°, equal to cos75°, and comes from using a minus sign in the addition formula.

cos105° = cos(60°+45°) = cos60°cos45°−sin60°sin45° = (√2/4)−(√6/4) = (√2−√6)/4, a negative value. The addition formula for cos is cos cos minus sin sin. The option (√6−√2)/4 reverses the sign, and you can check the answer by noting that 105° is obtuse, so its cosine must be negative.

tan(α+β) = (tanα+tanβ)/(1−tanαtanβ) = (2+3)/(1−6) = 5/(−5) = −1. The option 5/7 has the sign error of writing the denominator as 1+tanαtanβ. The denominator subtracts the product from 1.

The double angle formula gives sin2α = 2sinαcosα = 2×(3/5)×(4/5) = 24/25. The option 12/25 is the classic slip of forgetting the factor 2, and 7/25 is the value of cos2α = cos²α−sin²α.

cos2α = 1−2sin²α = 1−2×(1/9) = 7/9. When only sin is known, choose the form 1−2sin²α. The option −7/9 reverses the sign as 2sin²α−1, and 8/9 forgets the factor 2 and computes 1−1/9.

Substitute α=π/4 into sin²(α/2) = (1−cosα)/2: sin²(π/8) = (1−cos(π/4))/2 = (1−√2/2)/2 = (2−√2)/4. The option (2+√2)/4 is cos²(π/8) and comes from mixing up which of sin² and cos² takes the 1−cos form; it is sin².

√(1²+(√3)²) = √4 = 2, and the point (1,√3) has argument π/3, so the result is 2sin(θ+π/3). Substituting θ=0 gives √3 on the left and 2sin(π/3)=√3 on the right, which agree. The option 2sin(θ+π/6) confuses the point with (√3,1), and 4sin(…) forgets the square root of a²+b²=4.

Here a=1 and b=−1, so √(1+1)=√2 and the point (1,−1) has argument −π/4, giving √2 sin(θ−π/4). The option √2 sin(θ+π/4) is the combination of sinθ+cosθ and arises from overlooking the sign of b. Checking at θ=0 gives −1 on the left and √2 sin(−π/4)=−1 on the right.

Combining gives y = √(3²+4²) sin(θ+α) = 5sin(θ+α). Since the maximum of sin is 1, the maximum value is 5. The option 7 adds the coefficients as 3+4, and 25 forgets the square root of a²+b².

Since the angles are acute, cosα = 3/5 and sinβ = 12/13. Then cos(α+β) = cosαcosβ−sinαsinβ = 15/65−48/65 = −33/65. The option 63/65 uses a plus sign, and 56/65 is the value of sin(α+β). The result is negative because α+β turns out to be obtuse.

On the unit circle the y-coordinate equals 1/2 at one angle in the first quadrant and one in the second, namely θ = π/6 and π−π/6 = 5π/6. The values π/3, 2π/3 belong to sinθ=√3/2, and 7π/6 is a point with y=−1/2.

On the unit circle the x-coordinate equals −1/2 in the second and third quadrants, at θ = π−π/3 = 2π/3 and π+π/3 = 4π/3. The values π/3, 5π/3 solve cosθ=+1/2, missing the sign, and 5π/6, 7π/6 solve cosθ=−√3/2.

Since tan is a slope, a line of slope 1 meets the unit circle at two points symmetric about the origin: θ = π/4 and π/4+π = 5π/4. Solutions of a tangent equation are spaced π apart. The values 3π/4 and 7π/4 solve tanθ=−1.

The boundaries are the solutions of sinθ=1/2, namely π/6 and 5π/6. The y-coordinate exceeds 1/2 on the upper arc between those two solutions, so the answer is π/6 < θ < 5π/6. With no equality the endpoints are excluded. Read it off the unit circle as the part above the horizontal line y=1/2.

The boundaries are the solutions of cosθ=1/2, namely π/3 and 5π/3. The x-coordinate is at most 1/2 on the arc to the left of the vertical line x=1/2, so the answer is π/3 ≦ θ ≦ 5π/3. Since equality is allowed, both endpoints are included. Note that the range is a single arc straddling π.

Putting sinθ=t gives 2t²−t−1 = (2t+1)(t−1) = 0, so t = 1 or −1/2. From sinθ=1 we get θ=π/2, and from sinθ=−1/2 we get θ=7π/6 and 11π/6. List all three. Answering π/2 only drops the t=−1/2 branch.

Putting t=sinθ with −1≦t≦1 gives y = t²+t = (t+1/2)²−1/4. The vertex t=−1/2 lies inside the range, so the minimum is −1/4. The value 0 occurs at the ends t=−1 and t=0, and 2 is the maximum at t=1. Checking whether the vertex lies inside the range is the key step after a substitution.

Since −1 ≦ sinθ ≦ 1, the term 2sinθ runs from −2 to 2, and adding 1 gives −1 ≦ y ≦ 3. So the maximum is 3 and the minimum is −1. The pair maximum 2, minimum −2 forgets the shift by +1.

Combining gives √2 sin(θ+π/4) = 1, so sin(θ+π/4) = 1/√2. Over the range π/4 ≦ θ+π/4 < 9π/4 we get θ+π/4 = π/4 or 3π/4, that is θ = 0 or π/2. Checking, θ=0 gives 0+1=1 and θ=π/2 gives 1+0=1.

Combining gives y = 2sin(θ+π/3). As θ+π/3 runs from π/3 to 5π/6, the sine reaches its largest value 1 at θ+π/3=π/2 and its smallest value 1/2 at θ+π/3=5π/6, so the maximum is 2 and the minimum is 1. The minimum −2 ignores the restricted range, and the minimum √3 jumps to the endpoint value 2sin(π/3)=√3 at θ=0, whereas the value 1 at the θ=π/2 end is smaller.