Cos a-b formula

I understand how this video proves the angle addition for sine, but not where this formula comes from to begin with, I feel like somewhere I missed a step. It seems like a very complex proof for such a simple concept, why can't we just add sine a + sine b directly? What is it about trig functions that makes angle additions so complicated? I felt like I was grasping all the trig identities/unit circle definitions, etc up to this point but just crashed & burned here... The video is very clear but it seems like there should be some sort of introductory video to the concept of adding angles & why we can't do it more easily... Maybe it's just me? This is a good question. Here's a proof I just came up with that the angle addition formula for sin() applies to angles in the second quadrant: Given: pi/2 < a < pi and pi/2 < b < pi // a and b are obtuse angles less than 180°. Define: c = a - pi/2 and d = b - pi/2 // c and d are acute angles. Theorem: sin(c + d) = sin(c)*cos(d) + cos(c)*sin(d) // angle addition formula for sin(). Substitute: sin((a - pi/2) + (b - pi/2)) = sin(a - pi/2)*cos(b - pi/2) + cos(a - pi/2)*sin(b - pi/2) Simplify: sin((a - pi/2) + (b - pi/2)) = sin(a + b - pi) sin(a + b - pi) = -sin(a + b) // from unit circle sin(a - pi/2) = -cos(a) and sin(b - pi/2) = -cos(b) // from unit circle cos(a - pi/2) = sin(a) and cos(b - pi/2) = sin(b) // from unit circle Substitute: -sin(a + b) = -cos(a)*sin(b) + sin(a)*(-cos(b)) Simplify and rearrange: sin(a + b) = sin(a)*cos(b) + ...

The law of cosines calculator can help you solve a vast number of triangular problems. You will learn what is the law of cosines (also known as the cosine rule), the law of cosines formula, and its applications. Scroll down to find out when and how to use the law of cosines, and check out the proofs of this law. Thanks to this triangle calculator, you will be able to find the properties of any arbitrary triangle quickly. The law of cosines states that, for a triangle with sides and angles denoted with symbols as illustrated above, a² = b² + c² - 2bc × cos(α) b² = a² + c² - 2ac × cos(β) c² = a² + b² - 2ab × cos(γ) For a right triangle, the angle gamma, which is the angle between legs a and b, is equal to 90°. The cosine of 90° = 0, so in that special case, the law of cosines formula is reduced to the well-known equation of a² = b² + c² - 2bc × cos(90°) a² = b² + c² The law of cosines (alternatively the cosine formula or cosine rule) describes the relationship between the lengths of a triangle's sides and the cosine of its angles. It can be applied to all triangles, not only the right triangles. This law generalizes the Pythagorean theorem, as it allows you to calculate the length of one of the sides, given you know the length of both the other sides and the angle between them. AB² = CA² + CB² - 2 × CA × CH (for acute angles, '+' for obtuse) However, we may reformulate Euclid's theorem easily to the current cosine formula form: CH = CB × cos(γ), so AB² = CA² + CB² - 2 × CA ×...

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Cos(a + b) In trigonometry, cos(a + b) is one of the important trigonometric identities involving compound angle. It is one of the trigonometry formulas and is used to find the value of the cosine trigonometric function for the sum of angles. cos (a + b) is equal to cos a cos b - sin a sin b. This expansion helps in representing the value of cos trig function of a compound angle in terms of sine and cosine trigonometric functions. Let us understand the cos(a+b) identity and its proof in detail in the following sections. 1. 2. 3. 4. 5. Proof of Cos(a + b) Formula The verification of the expansion of cos(a+b) formula can be done geometrically. Let us see the stepwise derivation of the formula for the cosine To prove: cos (a + b) = cos a cos b - sin a sin b Construction: Assume a rotating line OX and let us rotate it about O in the anti-clockwise direction till it reaches Y. OX makes out an acute angle with Y given as, ∠XOY = a, from starting position to its final position. Again, this line rotates further in the same direction and starting from the position OY till it reaches Z, thus making out an acute angle given as, ∠YOZ = b. ∠XOZ = a + b < 90°. On the bounding line of the compound angle (a + b) take a point P on OZ, and draw PQ and PR Now, from the right-angled triangle PQO we get, cos (a + b) = OQ/OP = (OS - QS)/OP = OS/OP - QS/OP = OS/OP - TR/OP = OS/OR ∙ OR/OP + TR/PR ∙ PR/OP = cos a cos b - sin ∠TPR sin b = cos a cos b - sin a sin b, (since we know, ∠TPR = a) Therefo...

No, and there's a precise reason. First, the geometric definition of $\cos$ talks about angles, and the product of two angles doesn't make sense. Moreover, when you view the cosine as an exponential complex function, as you know $$\cos$, there isn't one for the $\cos$ function too. $\begingroup$ Dimensionally speaking, an angle is a dimensionless quantity; there's no reason that the product of two angles necessarily fails to make sense. (The second half of your answer actually illustrates why this is necessary, because you can't exponentiate dimensioned quantities; if angles weren't dimensionless, then the expression $e^$ wouldn't make dimensional sense. $\endgroup$ For general $a$ and $b$, we cannot write $\cos (ab)$ in terms of the trig functions $\cos a,\sin a, \cos b, \sin b$. This is because the trig functions are periodic with period $2\pi$, so adding $2\pi$ to $b$ does not change any of these functions. But adding $2\pi$ to $b$ can change $\cos (ab)$ - for instance, if $a=1/2$, if sends $\cos (ab)$ to $-\cos(ab)$. Only if $a$ is an integer can we avoid this problem. As many experts already noted here, an argument of cos(⋯) is an angle, and a sensible mathematical structure on angles is the one of There is no “reasonable” multiplication of an angle and an irrational number $t$. If $a$ is specified modulo 2π radians (that is a typical condition), then possible values of $ta$ will form a dense subset of the trigonometric circle, and hence values of trigonometric functi...

In this post, we will establish the formula of sin(a+b) sin(a-b). Note that sin(a+b) sin(a-b) is a product of two cosine functions. We will use the following two formulas: cos(a+b) = cos a cos b – sin a sin b …(i) cos(a-b) = cos a cos b + sin a sin b …(ii) Formula of cos(a+b) cos(a-b) cos(a+b) cos(a-b) Formula: cos(a+b) cos(a-b) = $\cos^2 a -\sin^2 b$ = $\cos^2 b -\sin^2 a$ Proof: Using the above formulas (i) and (ii), we have cos(a+b) cos(a-b) = (cos a cos b – sin a sin b) (cos a cos b + sin a sin b) = $(\cos a \cos b)^2$ $-(\sin a \sin b)^2$ Here we have used the formula (x+y)(x-y)=x 2-y 2 = $\cos^2 a \cos^2 b -\sin^2 a \sin^2b$ $\cdots (\star)$ = $\cos^2 a (1-\sin^2 b)$ $-(1-\cos^2 a) \sin^2b$ by the formula $sin^2 \theta+\cos^2\theta=1$ = $\cos^2 a – \cos^2 a \sin^2 b$ $-\sin^2b+\cos^2 a \sin^2b$ = cos2a -sin2b So the formula of cos(a+b) cos(a-b) is cos2a -sin2b. Next, we will prove that cos(a+b) cos(a-b) = cos2b -sin2a. Proof: From $(\star)$ we have that cos(a+b) cos(a-b) = $\cos^2 a \cos^2 b -\sin^2 a \sin^2b$ = $(1-\sin^2 a) \cos^2 b$ $-\sin^2 a (1-\cos^2b)$ = $\cos^2 b – \sin^2 a \cos^2 b$ $-\sin^2 a+ \sin^2 a \cos^2b$ = cos2b -sin2a (Proved) Also Read: In a similar way as above, we can prove the formula of cos(α+β) cos(α-β) = cos2α -sin2β. Formula of cos(α+β) cos(α-β) Prove that cos(α+β) cos(α-β) = cos2α -sin2β. Proof: Using the above formulas (i) and (ii), we have sin(α+β) sin(α-β) = (cos α cos β – sin α sin β) (cos α cos β + sin α sin β) = $(\cos \alpha \cos \be...

\( \newcommand\) • • • • • • Focus Questions The following questions are meant to guide our study of the material in this section. After studying this section, we should understand the concepts motivated by these questions and be able to write precise, coherent answers to these questions. • What are the Cosine Difference and Sum Identities? • What are the Sine Difference and Sum Identities? • What are the Tangent Difference and Sum Identities? • What are the Cofunction Identities? • Why are the difference and sum identities useful? The next identities we will investigate are the sum and difference identities for the cosine and sine. These identities will help us find exact values for the trigonometric functions at many more angles and also provide a means to derive even more identities. Beginning Activity • Is \(\cos(A - B) = \cos(A) - \cos(B)\) an identity? Explain. • Is \(\sin(A - B) = \sin(A) - \sin(B)\) an identity? Explain. • Use a graphing utility to draw the graph of \(y = \sin(\dfrac - x) = \cos(x)\) is an identity? Why or why not? The Cosine Difference Identity To this point we know the exact values of the trigonometric functions at only a few angles. Trigonometric identities can help us extend this list of angles at which we know exact values of the trigonometric functions. Consider, for example, the problem of finding the exact value of \(\cos(\dfrac)\). In our Beginning Activity, however, we saw that the equation \(\cos(A - B) = \cos(A) - \cos(B)\) is not an i...

The six trigonometric functions are defined for every sin ⁡ θ = o p p o s i t e h y p o t e n u s e = a h Pythagorean identities [ ] Main article: Identity 1: sin 2 ⁡ θ + cos 2 ⁡ θ = 1 Angle sum identities [ ] Draw a horizontal line (the x-axis); mark an origin O. Draw a line from O at an angle α Inequalities [ ] The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2 π) of the whole circle, so its area is θ/2. We assume here that θ π/2, then θ> 1. But sin θ ≤ 1 (because of the Pythagorean identity), so sin θ 1 if 0 1\quad See also [ ]

Cos(a + b) In trigonometry, cos(a + b) is one of the important trigonometric identities involving compound angle. It is one of the trigonometry formulas and is used to find the value of the cosine trigonometric function for the sum of angles. cos (a + b) is equal to cos a cos b - sin a sin b. This expansion helps in representing the value of cos trig function of a compound angle in terms of sine and cosine trigonometric functions. Let us understand the cos(a+b) identity and its proof in detail in the following sections. 1. 2. 3. 4. 5. Proof of Cos(a + b) Formula The verification of the expansion of cos(a+b) formula can be done geometrically. Let us see the stepwise derivation of the formula for the cosine To prove: cos (a + b) = cos a cos b - sin a sin b Construction: Assume a rotating line OX and let us rotate it about O in the anti-clockwise direction till it reaches Y. OX makes out an acute angle with Y given as, ∠XOY = a, from starting position to its final position. Again, this line rotates further in the same direction and starting from the position OY till it reaches Z, thus making out an acute angle given as, ∠YOZ = b. ∠XOZ = a + b < 90°. On the bounding line of the compound angle (a + b) take a point P on OZ, and draw PQ and PR Now, from the right-angled triangle PQO we get, cos (a + b) = OQ/OP = (OS - QS)/OP = OS/OP - QS/OP = OS/OP - TR/OP = OS/OR ∙ OR/OP + TR/PR ∙ PR/OP = cos a cos b - sin ∠TPR sin b = cos a cos b - sin a sin b, (since we know, ∠TPR = a) Therefo...

The six trigonometric functions are defined for every sin ⁡ θ = o p p o s i t e h y p o t e n u s e = a h Pythagorean identities [ ] Main article: Identity 1: sin 2 ⁡ θ + cos 2 ⁡ θ = 1 Angle sum identities [ ] Draw a horizontal line (the x-axis); mark an origin O. Draw a line from O at an angle α Inequalities [ ] The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2 π) of the whole circle, so its area is θ/2. We assume here that θ π/2, then θ> 1. But sin θ ≤ 1 (because of the Pythagorean identity), so sin θ 1 if 0 1\quad See also [ ]