Tutor profile: Varun S.
What is the value of $$cos(24°) + cos(5°) + cos(175°) + cos(204°) + cos(300°)$$?
Here is have multiple terms with the trigonometric function of cos. It is not possible to remember the value of cos for all these angles. So we need to simplify the terms, to arrive at a typical (more common ) angle like 30 or 45 or 60 degree. The most common method is to use periodic properties of trig functions. cos (180° - θ) = - cos θ cos (180° + θ) = - cos θ cos (360° - θ) = cos θ Lets use these properties to simplify the expression 1) cos(175°) = cos(180° - 5°) Using cos (180° - θ) = - cos θ cos(175°) = -cos(5°) 2) Using cos (180° + θ) = - cos θ cos(204°) =cos(180+24) = -cos(24) 3) Using cos (360° - θ) = cos θ cos(300°) = cos(360-60) = cos(60) Now substituting 1,2,3 back into the given expression, $$cos(24°) + cos(5°) -cos(5°) -cos(24) + cos(60)$$ This simplifies to just cos(60) We know the standard angle value for cos(60) = 1/2. So the answer is 1/2
Consider the Cartesian plane and the region represented the the inequality $$ 2x-3y <= -6 $$. Which quadrant of the plane will never be part of this region?
We have been given a linear inequality, and been asked a question about the region defined by it. So first we should find out the associated region. Let's simplify the expression. $$ 2x-3y <= -6 $$. We should always try to bring terms on y variable to the left hand side of the equation, and take x terms to the right hand side. This allows us to think about the inequality as a line. Taking 2x to the right hand side, => $$-3y <= -6 - 2x$$ Dividing both side by -3, and changing the inequality sign since we are dividing by a negative number, $$ y => 2 + 2x/3$$ We need to define the area under this inequality. Lets first look at the boundary or the edge. This is defined by the line $$ y = 2 + 2x/3$$ When x = 0, y = . So one point on the line is (0,2) on the y-axis When y=0, x = -3. So second point on the line is (-3,0) on the x-axis. With these two points, we can draw the line. This line does not cross quadrant 4. Now the region defined by $$ y => 2 + 2x/3$$ will always be above the line $$ y = 2 + 2x/3$$. So the region defined $$ y => 2 + 2x/3$$ will never include quadrant 4.
The variables x and y are related by the following equations - $$3^y - 1 = 2^x$$ and $$3^y = 11 - 2^(4-x)$$ If this system of equations has two sets of roots as (a,b) and (c,d), what is the value of a+b+c+d?
We have a system of 2 equations with variables as x and y. We have been told that there are 2 solutions to this system. So, first we should try to find the solutions. Equation 1 : $$3^y - 1 = 2^x$$ Equation 2 : $$3^y = 11 - 2^(4-x)$$ Looking at the structure, we can use a substitution from equation 1 into equation 2 for one of the variables. In this case, $$3^y$$ is common to both equations, so we can substitute $$3^y$$ from Equation 1 into equation 2 From Equation 1, $$3^y = 2^x + 1$$ Substituting this value of $$3^y$$ into equation 2, we get $$2^x + 1 + 2^(4-x) = 11$$ => $$2^x + 2^(4-x) = 10$$ Look at the part $$2^(4-x)$$. This term has a subtraction term in the power. We know that:- $$x^(a-b) = x^a/x^b$$. Using this formula, we can simplify the equation as:- $$2^x + 2^4/2^x = 10$$ Multipliying 2^x on both sides - $$2^(2x) + 16 = 10. 2^x$$ Now we define another variable z as $$z = 2^x $$ and substitute in the equation above. => $$z^2 - 10z + 16 = 0$$ This is a simple quadratic equation that can be solved by splitting the middle term. => $$z^2-8z-2z+16=0$$ =>$$z(z-8) - 2(z-8) = 0$$ => $$(z-2)(z-8)=0$$ Hence z = 2,8 Now coming back to the definition of z i.e. $$z = 2^x $$ If z=2, then $$2 = 2^x $$, so x=1 If z=8, then $$8 = 2^x $$, so x=3 We now have the 2 solution for x as 1 and 3. To get the corresponding value of y, we simply substitute x in equation 1 Equation 1 : $$3^y - 1 = 2^x$$ When x = 1, $$3^y - 1 = 2^1$$ => $$3^y = 3$$ => y=1 When x = 3, $$3^y - 1 = 2^3$$ => $$3^y = 9$$ => y=2 So the solutions to the system of equations are (1,1) and (3,2) That means a+b+c+d = 1+1+3+2 = 7. So the answer to this question is 7.
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