How do I find the value for the following trigonometric function?
Zero is the answer..a for alphaAnd b for theetatheSin(b+2a)÷sinb=1/3Componendo and dividendo{Sin(b+2a)-sinb}/{sin(b+2a)-sinb}=(1–3)/1+3=-2/4=-1/2On using formulas for sin(a+b)+sin(a-b) and sin(a+b) - sin(a-b)We ll get cot(b+2a)tan(a)=-1/22Tana+tan(b+2a)=0
How do I find the maximum value for the following trigonometric function?
There is a formula[math]-\sqrt{a^2+b^2} <= aSinx + bCosx <= \sqrt{a^2+b^2}[/math]so maximum value of[math]2Sinx + 4Cosx <= \sqrt{2^2+4^2}[/math]Add 3 both sides[math]2Sinx + 4Cosx +3 <= \sqrt{20} + 3[/math][math]2Sinx + 4Cosx +3 <= 2\sqrt{5} + 3[/math]So option (a) is correct.
What is the definition for principal value of inverse trigonometric functions?
When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each x in the domain the expression arcsin(x) will evaluate only to a single value, called its principal value.This is the definition for principal value of inverse trigonometry.
What is the value of the trigonometric function sin765°?
A positive angle is measured as a rotation from the “x axis” in an anticlockwise direction.This angle is 765 degrees!This means when you have rotated 360 degrees you keep on rotating!In this case you go right round twice and a bit more as shown below…This means that after a rotation of 765 degrees we are in the same POSITION as a 45 degree rotation!This means that sin 765 is the same as sin 45 = 1/√2 or ≈ 0.7071
What does it mean by principal values in trigonometry?
When you take the inverse trig function of a positive number the principal value is the answer in the first quadrant. When you take the inverse trig function of a negative number the principal value is the answer in either in the second or the fourth quadrant with the following rules: arcsin and arctan and arccsc are in the fourth quadrant. arccos and arccot and arcsec are in the second quadrant. If you aren't aware of it arcsin(x) means inverse sin(x) and it's the same for the other five trig functions. .
What are the principal values of trigonometry?
Lots of students have problems memorising trigonometric identities. I can give some rules by dint of which you will be able to solve such problems.If in the trigonometric function you have π/2+/-t, 3π+/-t, then you should change the name of the function to the reverse (sin-cos, tan-cot).If you have π+/-t, 2π+/-t, then you preserve the name of the function.Then you should place the sign before the function you have got that you would have on the condition that 0＜t＜π/2I know it is a bit vague, but let me give you examples so that make my claims more understandable.Let’s transform cos(π/2 +t). We change the name of the function on the reverse, as we have π/2. Now we have cos(π/2 +t)＝ _sin(t). Let’s find the sign. We have the stipulation that 0＜t＜π/2. We take, for example, π/4. Then we place it in the original function and get cos(π/2 +π/4). We can now deduce that the cos function is located in the second quadrant, where it has negative sign. Consequently, cos(π/2 +t)＝ -sin(t).If you want me to clarify you sth, write me in the comments.I hope this rule helps you to understand the trigonometric identities and solve such equations more easily.
How are the 'degree angle' of trigonometric functions related to its values.?
Draw a right angled isosceles triangle in which the angles are 45, 45 and 90. The sides will be in the form of a, a, a*sqrt (2). Now consider Sin (45). We get 1/ sqrt (2). Similarly consider an equilateral triangle and draw an altitude which is same as an angular bisector. Now there are two small right angled triangles whose angles are 30, 60 and 90. Now the sides will be in the form of a/2, a*sqrt(3)/2, a. Now consider sin (30) and sin (60).
How do I find the values of coordinates (probably using trigonometric functions)? The link is the question.
Any point on terminal side of angle [math]\alpha[/math] (in standard position) located on a circle with radius [math]r[/math], has coordinates: [math](r \cos\alpha, r\sin\alpha)[/math]The [math]4[/math] points shown are located on a circle with radius [math]8[/math]The [math]4[/math] angles are obtaining by dividing a right angle into [math]4[/math] equal angles.So the 4 angles are: [math]22.5^\circ[/math], [math]45^\circ[/math], [math]67.5^\circ[/math], [math]90^\circ[/math].We know the following:[math]\begin{array}{ll} \cos(45^\circ) = \sqrt2/2 & \cos(90^\circ) = 0 \\ \sin(45^\circ) = \sqrt2/2 & \cos(90^\circ) = 1 \end{array}[/math]Use half-angle identities to find sine and cosine of [math]\boldsymbol{22.5^\circ}[/math][math]\begin{align*} \cos^2(22.5^\circ) & = \dfrac{1+\cos(45^\circ)}{2} \\ & = \frac{1+\frac{\sqrt 2}{2}}{2} \\ & = \frac{2+\sqrt 2}{4} \\ \cos(22.5^\circ) & =\frac{\sqrt{2+\sqrt{2}}}{2} \\ \end{align*}[/math][math]\begin{align*} \sin^2(22.5^\circ) & = \dfrac{1-\cos(45^\circ)}{2} \\ & = \frac{1-\frac{\sqrt 2}{2}}{2} \\ & = \frac{2-\sqrt 2}{4} \\ \sin(22.5^\circ) & =\frac{\sqrt{2-\sqrt{2}}}{2} \end{align*}[/math]Use co-functions to find sine and cosine of [math]\boldsymbol{67.5^\circ}[/math][math]\cos(67.5^\circ) = \sin(90^\circ\!-\!67.5^\circ) = \sin(22.5^\circ) = \frac{\sqrt{2-\sqrt{2}}}{2} \\ \sin(67.5^\circ) = \cos(90^\circ\!-\!67.5^\circ) = \cos(22.5^\circ) = \frac{\sqrt{2+\sqrt{2}}}{2}[/math]Coordinates:[math](x_1,y_1)= (8 \cos(22.5^\circ), 8 \sin(22.5^\circ)) = \left( 4\sqrt{2+\sqrt{2}}, 4\sqrt{2-\sqrt{2}} \right)[/math][math](x_2,y_2)= (8 \cos(45^\circ), 8 \sin(45^\circ)) = \left( 4\sqrt{2}, 4\sqrt{2} \right)[/math][math](x_3,y_3)= (8 \cos(67.5^\circ), 8 \sin(67.5^\circ)) = \left( 4\sqrt{2-\sqrt{2}}, 4\sqrt{2+\sqrt{2}} \right)[/math][math](x_4,y_4)= (8 \cos(90^\circ), 8 \sin(90^\circ)) = \left( 0, 8 \right)[/math]
Finding the principal value in complex analysis?
1) 4^(1/2) = 2. 2) i^(2i) = exp(2i * Log i), via principal value = exp(2i * (ln |i| + i Arg i)) = exp(2i * (ln(1) + iπ/2)) = exp(-π). 3) (1+i)^(1+i) = exp((1+i) * Log(1+i)), via principal value = exp((1+i) * (ln |1+i| + i Arg(1+i))) = exp((1+i) * (ln√2 + iπ/4)) = exp((ln√2 - π/4) + i(π/4 + ln√2)) = exp(ln√2 - π/4) [cos(π/4 + ln√2) + i sin(π/4 + ln√2)]. I hope this helps!