Find exact answer for each expression(TRIG)?
Careful! I hope that was just a typo in the first problem. You meant 8pi + pi/4, I'm sure. The period of cosine is 2*pi. So, for any integer n cos(x+n*2pi) = cos(x). cos(33pi/4) = cos(4*2pi + pi/4) = cos(pi/4). The period of the tangent function is 1 pi not 2 pi. So, for any integer n tan(x+n*pi) is the same as tan(x) tan(21pi) = tan(0 + 21pi) = tan(0) = 0
Trig. Functions: finding exact solutions?
1. solve exactly if possible or to four significant digits if an exact solution can be found. (show all work) 7cos²x = 8sinx + 8 with 0≤ x ≤ 2π 2. Find all exact solutions to (show all work) sinx + 1 = cosx with 0≤ x ≤ 2π 3. Find all exact real solutions to (show all work) 4sinx + 2 =0 with 0≤ x ≤ 2π thanks you all..
How to find exact trig functions?
It is past your test, but in case you still need to know... 1035 degrees is coterminal (ends in the same spot) with 315 degrees (because 360 + 360 + 315 = 1035...twice around the circle plus 315 more degrees). 315 is one of the "standard points" on a unit circle, with a cosine value of 1/sqrt(2), a sine value of -1/sqrt(2), and a tangent value of -1. Cosecant is the reciprocal of sine, so take the reciprocal of -1/sqrt(2)...answer: -sqrt(2). 35pi/6 is coterminal with 5pi/6 (35pi/6 = 6pi + 5pi/6, so three times around the circle plus 5pi/6 more). 5pi/6 has a cosine value of -sqrt(3)/2. Hope that helps some.
Finding exact trig values?
Let A = arcsin(5/13) sin A = 5/13 Since 5/13 > 0, we know 0° <= A <= 90° sin²A + cos²A = 1 (5/13)² + cos²A = 1 25/169 + cos²A = 1 cos²A = 144/169 cos A = 12/13 OR cos A = -12/13 Since 0° <= A <= 90°, cos A = 12/13 - - - - - Let B = arctan(-4/3) tan B = -4/3 Since -4/3 < 0, we know -90° <= B <= 0° sin²B + cos²B = 1 (sin²B/cos²B) + (cos²B/cos²B) = 1/cos²B tan²B + 1 = 1/cos²B (-4/3)² + 1 = 1/cos²B (16/9) + 1 = 1/cos²B 25/9 = 1/cos²B cos²B = 9/25 cos B = 3/5 OR cos B = -3/5 Since -90° <= B <= 0°, cos B = 3/5 sin²B + cos²B = 1 sin²B + (3/5)² = 1 sin²B + (9/25) = 1 sin²B = 16/25 sin B = 4/5 OR sin B = -4/5 Since -90° <= B <= 0°, sin B = -4/5 - - - - - cos( arcsin(5/13) + arctan(-4/3) ) = cos( A + B ) = cos A x cos B - sin A x sin B = (12/13) x (3/5) - (5/13) x (-4/5) = 56/65
Find the exact value of the trig function?
Hopefully you are familiar with the two 'integer value' right-angle triangles with sides 3-4-5 and 7-24-25, and with the formula for expanding cos(u + v), which is: cos(u + v) = cos(u)cos(v) - sin(u)sin(v). From diagrams of those two triangles you can read any of the trig ratios for the quadrant I angle which is the reference angle for your quadrant III angles. When the absolute value of the sine is 7/25, the corresponding cosine has an absolute value of 24/25. When the absolute value of the cosine is 4/5, the absolute value of the sine is 3/5. In quadrant III, sines and cosines are both negative. Therefore cos(u) = -24/25, and sin(v) = -3/5. Substituting in the formula: cos(u + v) = (-24/25)(-4/5) - (-7/25)(-3/5) = 24 * 4 / 125 - 7 * 3 / 125 = (96 - 21) / 125 = 75 / 125 = 3 / 5.
Finding exact trigonometric ratios?
yea u memorize them based on the exact trig triangles and the quadrant rules below i have links to the triangles and the quadrant rules. write them down and MEMORISE them!!! :D as pi = 180 (pi/180 = 1) to get from degree to radian: 45 deg = 45x pi/180 (so essentially multiplying by 1) = pi/4 you find that 3pi/4 = pi - pi/4 = 135 deg (180 - 45 = 135) - source 5 shows 135 in the 2nd quad pi minus any acute angle is in the 2nd quadrant (refer to third source) Ok, so you therefore have pi/4 is in the 2nd quad therefore, refering to source one, it is equal to 1/root2 as it is positive, it is for sin as sin is the only positive in the 2nd quad source 4 gives a really good overview of it all...look around on the different topics hope this helps :D
How do you find the exact trigonometric ratios for the angle 3pi/4?
With every angle there is associated a reference angle, which is the positive acute angle that the terminal side of the given angle makes with the (positive or negative) x-axis. An angle and its reference angle have the same value for their trig functions, except possibly for sign. 3π/4 is in the second quadrant, so its cosine is negative and its sine is positive. From these you can determine the sign of the other four trig functions. The reference angle for 3π/4 is π/4. This is an angle for which you are expected to know at least the sine and cosine: sin(π/4) = cos(π/4) = √2 / 2. Therefore, sin(3π/4) = √2 / 2 cos(3π/4) = -√2 / 2 tan(3π/4) = sin(3π/4)/cos(3π/4) = -1 csc(3π/4) = 1/sin(3π/4) = √2 sec(3π/4) = 1/cos(3π/4) = -√2 cot(3π/4) = 1/tan(3π/4) = -1
Find Exact Values of Trig Functions?
a) This is already in its most reduced form. b) sin(x - pi/6) Use the sine subtraction identity. sin(a - b) = sin(a)cos(b) - sin(b)cos(a) sin(x - pi/6) = sin(x)cos(pi/6) - sin(pi/6)cos(x) = sin(x)(sqrt(3)/2) - (1/2)cos(x) = (1/2) ( sqrt(3) sin(x) - cos(x) ) c) cos(a + b) = cos(a)cos(b) - sin(a)sin(b) cos(x + pi/3) = cos(x)cos(pi/3) - sin(x)sin(pi/3) = cos(x)(1/2) - sin(x)sqrt(3)/2 = (1/2) ( cos(x) - sqrt(3) sin(x) ) d) tan(a - b) = [ tan(a) - tan(b) ] / [ 1 + tan(a)tan(b)] tan(x - pi/4) = [ tan(x) - tan(pi/4) ] / [ 1 + tan(x)tan(pi/4) ] = [ tan(x) - 1 ] / [ 1 + tan(x)(1) ] = [ tan(x) - 1 ] / [ tan(x) + 1 ]
How can I get an exact value for inverse trig functions?
Sorry. What you're asking for isn't possible. When you take the inverse tangent of a decimal number in your calculator, your calculator doesn't actually know infinitely many digits of that number. It's a real-world thing, so it has to store only finitely many digits in its memory. Because of this, the result that you're calculator gives is not exact. It's an approximation. It's important to realize the tremendous difference between an approximate decimal, and an expression like [math]\frac{5\pi}{6}[/math], which is a number that is perfectly, exactly the number that you want, not just a cheap approximation of it.The answer to your question about getting an exact value is that you need to learn how to do trigonometry by hand. For instance, I can tell that the argument of the arctan function in your example probably is supposed to be $\frac{1}{\sqrt{3}}$. When you do calculations on paper, keep your numbers expressed exactly with roots rather than writing decimal forms of numbers, and manipulate the algebra this way, that way when you reach the end of the calculation the answer will not be just an approximation.
Trig questions? (ferris wheel, finding exact values etc)?
1. Ferris wheel question; h(t)=11-10cos(0.1πt) time in seconds. How long will it take a passenger to reach a height of 5 m fo the first time? Answer to 3 decimal places. 2. solve 1- tan^2x as exact sol'n. 0≤ x ≤ 2π; 3. If sinθ = cosθ = 1/√2 determine the value of sin(2θ) Is it; 0? 1? 2? √2? √2/2 ? 4. Evaluate log2(sec1.2) Is it; 0.392? 1.465? 2.003? 3.909? 4.118? 5. helicopter blade's diameter is 8m. speed of sound is approx. 1152km/h. Maximum angular velocicy in rad/seconds and rotations per second so that the blade tips do not exceed the speed of sound. any help would be greatly appreciated thank you.