cos(α) = sin(90° - α)
→ cos(16° + θ) = sin(90° - (16° + θ)) = sin(74° - θ)
→ sin(36° + θ) = cos(16° + θ)
→ sin((36° + θ) = sin(74° - θ)
→ 36° + θ = 74° - θ
→ 2θ = 38°
→ θ = 19°
→ θ = 19 °+ 180°n for n= 0, 1, 2, ...
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
Cosine squared theta = 1 + Sine squared theta
2 sin(x) + 1 = 0 2 sin(x) = -1 sin(x) = -1/2 x = 210° and 330°
34
The reciprocal of a number plus the reciprocal of twice the number equals Find the number. 1/2 find the number
tan2(theta) + 5*tan(theta) = 0 => tan(theta)*[tan(theta) + 5] = 0=> tan(theta) = 0 or tan(theta) = -5If tan(theta) = 0 then tan(theta) + cot(theta) is not defined.If tan(theta) = -5 then tan(theta) + cot(theta) = -5 - 1/5 = -5.2
Yes, it is.
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
2 sin (Θ) + 1 = 0sin (Θ) = -1/2Θ = 210°Θ = 330°
Cosine squared theta = 1 + Sine squared theta
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
It is a trigonometric equation.
1
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
The question contains an expression but not an equation. An expression cannot be solved.
2 sin(x) + 1 = 0 2 sin(x) = -1 sin(x) = -1/2 x = 210° and 330°